discrete mathematics-mathematical induction
TRANSCRIPT
![Page 1: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/1.jpg)
Applied Discrete MathematicsMathematical Induction
William Shoaff
Spring 2008
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Outline
1 Introduction to Induction
2 Induction on Sequences
3 Induction on Sums
![Page 3: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/3.jpg)
Outline
1 Introduction to Induction
2 Induction on Sequences
3 Induction on Sums
![Page 4: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/4.jpg)
Outline
1 Introduction to Induction
2 Induction on Sequences
3 Induction on Sums
![Page 5: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/5.jpg)
Quotation
The concept of “mathematical induction”should be distinguished from what is usually called
“inductive reasoning” in science.– Don Knuth “The Art of Computer Programming: Fundamental
Algorithms”American computer scientist (1938 – )
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Mathematical Induction
Inductive reasoning is a logic used for discovery in science
Repeated observation of consistent results leads to a conjecture
The principle of mathematical induction is an axiom in thetheory of natural numbers
Mathematical induction is a template procedure used toestablish facts about enumerable setsThere is a basis for induction, an observation of the fact insome instanceAnd there is a material implication that “if the fact is true insome instance, then it will be true in the next instance”
![Page 7: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/7.jpg)
Mathematical Induction
Inductive reasoning is a logic used for discovery in science
Repeated observation of consistent results leads to a conjectureThe principle of mathematical induction is an axiom in thetheory of natural numbers
Mathematical induction is a template procedure used toestablish facts about enumerable setsThere is a basis for induction, an observation of the fact insome instanceAnd there is a material implication that “if the fact is true insome instance, then it will be true in the next instance”
![Page 8: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/8.jpg)
Mathematical Induction
Inductive reasoning is a logic used for discovery in scienceRepeated observation of consistent results leads to a conjecture
The principle of mathematical induction is an axiom in thetheory of natural numbers
Mathematical induction is a template procedure used toestablish facts about enumerable setsThere is a basis for induction, an observation of the fact insome instanceAnd there is a material implication that “if the fact is true insome instance, then it will be true in the next instance”
![Page 9: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/9.jpg)
Mathematical Induction
Inductive reasoning is a logic used for discovery in scienceRepeated observation of consistent results leads to a conjecture
The principle of mathematical induction is an axiom in thetheory of natural numbers
Mathematical induction is a template procedure used toestablish facts about enumerable setsThere is a basis for induction, an observation of the fact insome instanceAnd there is a material implication that “if the fact is true insome instance, then it will be true in the next instance”
![Page 10: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/10.jpg)
Mathematical Induction
Inductive reasoning is a logic used for discovery in scienceRepeated observation of consistent results leads to a conjecture
The principle of mathematical induction is an axiom in thetheory of natural numbers
Mathematical induction is a template procedure used toestablish facts about enumerable sets
There is a basis for induction, an observation of the fact insome instanceAnd there is a material implication that “if the fact is true insome instance, then it will be true in the next instance”
![Page 11: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/11.jpg)
Mathematical Induction
Inductive reasoning is a logic used for discovery in scienceRepeated observation of consistent results leads to a conjecture
The principle of mathematical induction is an axiom in thetheory of natural numbers
Mathematical induction is a template procedure used toestablish facts about enumerable setsThere is a basis for induction, an observation of the fact insome instance
And there is a material implication that “if the fact is true insome instance, then it will be true in the next instance”
![Page 12: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/12.jpg)
Mathematical Induction
Inductive reasoning is a logic used for discovery in scienceRepeated observation of consistent results leads to a conjecture
The principle of mathematical induction is an axiom in thetheory of natural numbers
Mathematical induction is a template procedure used toestablish facts about enumerable setsThere is a basis for induction, an observation of the fact insome instanceAnd there is a material implication that “if the fact is true insome instance, then it will be true in the next instance”
![Page 13: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/13.jpg)
Observations: Full Binary Trees
The full binary tree of height h = 0 has 1 node
The full binary tree of height h = 1 has 3 nodes
1
2 3
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Observations: Full Binary Trees
The full binary tree of height h = 0 has 1 node
The full binary tree of height h = 1 has 3 nodes
1
2 3
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Observations: Full Binary Trees
The full binary tree of height h = 0 has 1 node
The full binary tree of height h = 1 has 3 nodes
1
2 3
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Observations: Full Binary Trees
The full binary tree of height h = 2 has 7 nodes
1
2
4 5
3
6 7
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Observations: Full Binary Trees
The full binary tree of height h = 2 has 7 nodes
1
2
4 5
3
6 7
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Question
What is the functional relationship between the height h of afull binary tree and the number of nodes n in the tree?
ObservationsHeight Nodesh = 0 n = 1h = 1 n = 3h = 2 n = 7
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Question
What is the functional relationship between the height h of afull binary tree and the number of nodes n in the tree?
ObservationsHeight Nodesh = 0 n = 1h = 1 n = 3h = 2 n = 7
![Page 20: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/20.jpg)
Question
What is the functional relationship between the height h of afull binary tree and the number of nodes n in the tree?
ObservationsHeight Nodesh = 0 n = 1h = 1 n = 3h = 2 n = 7
![Page 21: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/21.jpg)
Question
What is the functional relationship between the height h of afull binary tree and the number of nodes n in the tree?
ObservationsHeight Nodesh = 0 n = 1h = 1 n = 3h = 2 n = 7
![Page 22: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/22.jpg)
Hypothesis
The full binary tree of height h has n = 2h+1 − 1 nodesThis statement is true for h = 0, 1 and 2
Height Nodes0 1 = 2(0+1) − 11 3 = 2(1+1) − 12 7 = 2(2+1) − 1
To prove the statement for every integer
show that if it is true for some height h ≥ 0, then it will betrue for height h + 1
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Hypothesis
The full binary tree of height h has n = 2h+1 − 1 nodes
This statement is true for h = 0, 1 and 2
Height Nodes0 1 = 2(0+1) − 11 3 = 2(1+1) − 12 7 = 2(2+1) − 1
To prove the statement for every integer
show that if it is true for some height h ≥ 0, then it will betrue for height h + 1
![Page 24: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/24.jpg)
Hypothesis
The full binary tree of height h has n = 2h+1 − 1 nodesThis statement is true for h = 0, 1 and 2
Height Nodes0 1 = 2(0+1) − 11 3 = 2(1+1) − 12 7 = 2(2+1) − 1
To prove the statement for every integer
show that if it is true for some height h ≥ 0, then it will betrue for height h + 1
![Page 25: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/25.jpg)
Hypothesis
The full binary tree of height h has n = 2h+1 − 1 nodesThis statement is true for h = 0, 1 and 2
Height Nodes0 1 = 2(0+1) − 11 3 = 2(1+1) − 12 7 = 2(2+1) − 1
To prove the statement for every integer
show that if it is true for some height h ≥ 0, then it will betrue for height h + 1
![Page 26: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/26.jpg)
Hypothesis
The full binary tree of height h has n = 2h+1 − 1 nodesThis statement is true for h = 0, 1 and 2
Height Nodes0 1 = 2(0+1) − 11 3 = 2(1+1) − 12 7 = 2(2+1) − 1
To prove the statement for every integer
show that if it is true for some height h ≥ 0, then it will betrue for height h + 1
![Page 27: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/27.jpg)
Hypothesis
The full binary tree of height h has n = 2h+1 − 1 nodesThis statement is true for h = 0, 1 and 2
Height Nodes0 1 = 2(0+1) − 11 3 = 2(1+1) − 12 7 = 2(2+1) − 1
To prove the statement for every integershow that if it is true for some height h ≥ 0, then it will betrue for height h + 1
![Page 28: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/28.jpg)
Height h, Full Binary Tree Has 2h+1 − 1 Nodes
A full binary tree of height h + 1 is constructed from
1 root nodeA full left subtree of of height hA full right subtree of of height h
root
leftsubtree
rightsubtree
height h
height 1
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Height h, Full Binary Tree Has 2h+1 − 1 Nodes
A full binary tree of height h + 1 is constructed from
1 root nodeA full left subtree of of height hA full right subtree of of height h
root
leftsubtree
rightsubtree
height h
height 1
![Page 30: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/30.jpg)
Height h, Full Binary Tree Has 2h+1 − 1 Nodes
A full binary tree of height h + 1 is constructed from1 root node
A full left subtree of of height hA full right subtree of of height h
root
leftsubtree
rightsubtree
height h
height 1
![Page 31: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/31.jpg)
Height h, Full Binary Tree Has 2h+1 − 1 Nodes
A full binary tree of height h + 1 is constructed from1 root nodeA full left subtree of of height h
A full right subtree of of height h
root
leftsubtree
rightsubtree
height h
height 1
![Page 32: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/32.jpg)
Height h, Full Binary Tree Has 2h+1 − 1 Nodes
A full binary tree of height h + 1 is constructed from1 root nodeA full left subtree of of height hA full right subtree of of height h
root
leftsubtree
rightsubtree
height h
height 1
![Page 33: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/33.jpg)
Height h, Full Binary Tree Has 2h+1 − 1 Nodes
A full binary tree of height h + 1 is constructed from1 root nodeA full left subtree of of height hA full right subtree of of height h
root
leftsubtree
rightsubtree
height h
height 1
![Page 34: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/34.jpg)
Height h, Full Binary Tree Has 2h+1 − 1 Nodes
A full binary tree of height h + 1 is constructed from1 root nodeA full left subtree of of height hA full right subtree of of height h
root
leftsubtree
rightsubtree
height h
height 1
![Page 35: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/35.jpg)
Height h, Full Binary Tree Has 2h+1 − 1 Nodes
root
leftsubtree
rightsubtree
height h
height 1
If the left and right subtree both have 2h+1 − 1 nodes, thenthe nodes in the higher tree are
(1 root node) + (nodes in left) + (nodes in right)
1 + (2h − 1) + (2h+1 − 1) = 1 + 2(2h+1 − 1)
= 1 + (2h+2 − 2)
= 2h+2 − 1
![Page 36: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/36.jpg)
Height h, Full Binary Tree Has 2h+1 − 1 Nodes
root
leftsubtree
rightsubtree
height h
height 1
If the left and right subtree both have 2h+1 − 1 nodes,
thenthe nodes in the higher tree are
(1 root node) + (nodes in left) + (nodes in right)
1 + (2h − 1) + (2h+1 − 1) = 1 + 2(2h+1 − 1)
= 1 + (2h+2 − 2)
= 2h+2 − 1
![Page 37: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/37.jpg)
Height h, Full Binary Tree Has 2h+1 − 1 Nodes
root
leftsubtree
rightsubtree
height h
height 1
If the left and right subtree both have 2h+1 − 1 nodes,
thenthe nodes in the higher tree are
(1 root node) + (nodes in left) + (nodes in right)
1 + (2h − 1) + (2h+1 − 1) = 1 + 2(2h+1 − 1)
= 1 + (2h+2 − 2)
= 2h+2 − 1
![Page 38: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/38.jpg)
Height h, Full Binary Tree Has 2h+1 − 1 Nodes
root
leftsubtree
rightsubtree
height h
height 1
If the left and right subtree both have 2h+1 − 1 nodes, thenthe nodes in the higher tree are
(1 root node) + (nodes in left) + (nodes in right)
1 + (2h − 1) + (2h+1 − 1) = 1 + 2(2h+1 − 1)
= 1 + (2h+2 − 2)
= 2h+2 − 1
![Page 39: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/39.jpg)
Height h, Full Binary Tree Has 2h+1 − 1 Nodes
root
leftsubtree
rightsubtree
height h
height 1
If the left and right subtree both have 2h+1 − 1 nodes, thenthe nodes in the higher tree are
(1 root node) + (nodes in left) + (nodes in right)
1 + (2h − 1) + (2h+1 − 1) = 1 + 2(2h+1 − 1)
= 1 + (2h+2 − 2)
= 2h+2 − 1
![Page 40: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/40.jpg)
Height h, Full Binary Tree Has 2h+1 − 1 Nodes
root
leftsubtree
rightsubtree
height h
height 1
If the left and right subtree both have 2h+1 − 1 nodes, thenthe nodes in the higher tree are
(1 root node) + (nodes in left) + (nodes in right)
1 + (2h − 1) + (2h+1 − 1) = 1 + 2(2h+1 − 1)
= 1 + (2h+2 − 2)
= 2h+2 − 1
![Page 41: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/41.jpg)
Height h, Full Binary Tree Has 2h+1 − 1 Nodes
root
leftsubtree
rightsubtree
height h
height 1
If the left and right subtree both have 2h+1 − 1 nodes, thenthe nodes in the higher tree are
(1 root node) + (nodes in left) + (nodes in right)
1 + (2h − 1) + (2h+1 − 1)
= 1 + 2(2h+1 − 1)
= 1 + (2h+2 − 2)
= 2h+2 − 1
![Page 42: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/42.jpg)
Height h, Full Binary Tree Has 2h+1 − 1 Nodes
root
leftsubtree
rightsubtree
height h
height 1
If the left and right subtree both have 2h+1 − 1 nodes, thenthe nodes in the higher tree are
(1 root node) + (nodes in left) + (nodes in right)
1 + (2h − 1) + (2h+1 − 1) = 1 + 2(2h+1 − 1)
= 1 + (2h+2 − 2)
= 2h+2 − 1
![Page 43: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/43.jpg)
Height h, Full Binary Tree Has 2h+1 − 1 Nodes
root
leftsubtree
rightsubtree
height h
height 1
If the left and right subtree both have 2h+1 − 1 nodes, thenthe nodes in the higher tree are
(1 root node) + (nodes in left) + (nodes in right)
1 + (2h − 1) + (2h+1 − 1) = 1 + 2(2h+1 − 1)
= 1 + (2h+2 − 2)
= 2h+2 − 1
![Page 44: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/44.jpg)
Height h, Full Binary Tree Has 2h+1 − 1 Nodes
root
leftsubtree
rightsubtree
height h
height 1
If the left and right subtree both have 2h+1 − 1 nodes, thenthe nodes in the higher tree are
(1 root node) + (nodes in left) + (nodes in right)
1 + (2h − 1) + (2h+1 − 1) = 1 + 2(2h+1 − 1)
= 1 + (2h+2 − 2)
= 2h+2 − 1
![Page 45: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/45.jpg)
Gauss Fools the Third Grade Teacher
There is an often told story, with many variations, of howyoung Gauss irritated his third grade teacherThe class was asked to compute the value of the sum
1 + 2 + 3 + · · ·+ 99 + 100
Almost immediately Gauss answered 5050Perhaps Gauss knew the total would be the number of terms(100) times the average of the first and last term
1+2+3+ · · ·+99+100 = 100(1 + 100
2
)= 50(101) = 5050
![Page 46: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/46.jpg)
Gauss Fools the Third Grade Teacher
There is an often told story, with many variations, of howyoung Gauss irritated his third grade teacher
The class was asked to compute the value of the sum
1 + 2 + 3 + · · ·+ 99 + 100
Almost immediately Gauss answered 5050Perhaps Gauss knew the total would be the number of terms(100) times the average of the first and last term
1+2+3+ · · ·+99+100 = 100(1 + 100
2
)= 50(101) = 5050
![Page 47: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/47.jpg)
Gauss Fools the Third Grade Teacher
There is an often told story, with many variations, of howyoung Gauss irritated his third grade teacherThe class was asked to compute the value of the sum
1 + 2 + 3 + · · ·+ 99 + 100
Almost immediately Gauss answered 5050Perhaps Gauss knew the total would be the number of terms(100) times the average of the first and last term
1+2+3+ · · ·+99+100 = 100(1 + 100
2
)= 50(101) = 5050
![Page 48: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/48.jpg)
Gauss Fools the Third Grade Teacher
There is an often told story, with many variations, of howyoung Gauss irritated his third grade teacherThe class was asked to compute the value of the sum
1 + 2 + 3 + · · ·+ 99 + 100
Almost immediately Gauss answered 5050
Perhaps Gauss knew the total would be the number of terms(100) times the average of the first and last term
1+2+3+ · · ·+99+100 = 100(1 + 100
2
)= 50(101) = 5050
![Page 49: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/49.jpg)
Gauss Fools the Third Grade Teacher
There is an often told story, with many variations, of howyoung Gauss irritated his third grade teacherThe class was asked to compute the value of the sum
1 + 2 + 3 + · · ·+ 99 + 100
Almost immediately Gauss answered 5050Perhaps Gauss knew the total would be the number of terms(100) times the average of the first and last term
1+2+3+ · · ·+99+100 = 100(1 + 100
2
)= 50(101) = 5050
![Page 50: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/50.jpg)
Generalizing the Problem
Consider the more general sum
0 + 1 + 2 + 3 + · · ·+ (n − 2) + (n − 1)
The sum has n termsThe first term is 0, the last term is (n − 1), and their averageis (n − 1)/2Is it true that
0 + 1 + 2 + 3 + · · ·+ (n − 2) + (n − 1) = n(n − 12
)?
![Page 51: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/51.jpg)
Generalizing the Problem
Consider the more general sum
0 + 1 + 2 + 3 + · · ·+ (n − 2) + (n − 1)
The sum has n termsThe first term is 0, the last term is (n − 1), and their averageis (n − 1)/2Is it true that
0 + 1 + 2 + 3 + · · ·+ (n − 2) + (n − 1) = n(n − 12
)?
![Page 52: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/52.jpg)
Generalizing the Problem
Consider the more general sum
0 + 1 + 2 + 3 + · · ·+ (n − 2) + (n − 1)
The sum has n terms
The first term is 0, the last term is (n − 1), and their averageis (n − 1)/2Is it true that
0 + 1 + 2 + 3 + · · ·+ (n − 2) + (n − 1) = n(n − 12
)?
![Page 53: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/53.jpg)
Generalizing the Problem
Consider the more general sum
0 + 1 + 2 + 3 + · · ·+ (n − 2) + (n − 1)
The sum has n termsThe first term is 0, the last term is (n − 1), and their averageis (n − 1)/2
Is it true that
0 + 1 + 2 + 3 + · · ·+ (n − 2) + (n − 1) = n(n − 12
)?
![Page 54: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/54.jpg)
Generalizing the Problem
Consider the more general sum
0 + 1 + 2 + 3 + · · ·+ (n − 2) + (n − 1)
The sum has n termsThe first term is 0, the last term is (n − 1), and their averageis (n − 1)/2Is it true that
0 + 1 + 2 + 3 + · · ·+ (n − 2) + (n − 1) = n(n − 12
)?
![Page 55: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/55.jpg)
Prove∑
k = n(n−1)2
To prove
0 + 1 + 2 + 3 + · · ·+ (n − 2) + (n − 1) = n(n − 12
)is true for all natural numbers n, show
Basis: For n = 0
The sum on the left is empty, and so equal to 0
0+ 1+ · · ·+(0− 2)+ (0− 1) = 0 (empty sum)
The right hand side is also equal to 0 for n = 0
0(0− 12
)= 0
![Page 56: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/56.jpg)
Prove∑
k = n(n−1)2
To prove
0 + 1 + 2 + 3 + · · ·+ (n − 2) + (n − 1) = n(n − 12
)is true for all natural numbers n, show
Basis: For n = 0
The sum on the left is empty, and so equal to 0
0+ 1+ · · ·+(0− 2)+ (0− 1) = 0 (empty sum)
The right hand side is also equal to 0 for n = 0
0(0− 12
)= 0
![Page 57: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/57.jpg)
Prove∑
k = n(n−1)2
To prove
0 + 1 + 2 + 3 + · · ·+ (n − 2) + (n − 1) = n(n − 12
)is true for all natural numbers n, show
Basis: For n = 0
The sum on the left is empty, and so equal to 0
0+ 1+ · · ·+(0− 2)+ (0− 1) = 0 (empty sum)
The right hand side is also equal to 0 for n = 0
0(0− 12
)= 0
![Page 58: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/58.jpg)
Prove∑
k = n(n−1)2
To prove
0 + 1 + 2 + 3 + · · ·+ (n − 2) + (n − 1) = n(n − 12
)is true for all natural numbers n, show
Basis: For n = 0The sum on the left is empty, and so equal to 0
0+ 1+ · · ·+(0− 2)+ (0− 1) = 0 (empty sum)
The right hand side is also equal to 0 for n = 0
0(0− 12
)= 0
![Page 59: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/59.jpg)
Prove∑
k = n(n−1)2
To prove
0 + 1 + 2 + 3 + · · ·+ (n − 2) + (n − 1) = n(n − 12
)is true for all natural numbers n, show
Basis: For n = 0The sum on the left is empty, and so equal to 0
0+ 1+ · · ·+(0− 2)+ (0− 1) = 0 (empty sum)
The right hand side is also equal to 0 for n = 0
0(0− 12
)= 0
![Page 60: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/60.jpg)
Prove∑
k = n(n−1)2
Induction: Pretend
0 + 1 + 2 + · · ·+ (n − 2) + (n − 1) =n(n − 1)
2
for some n ≥ 0Then the next sum is:
[0 + 1 + · · ·+ (n − 1)] + n =n(n − 1)
2+ n
=n(n − 1)
2+
2n2
=n(n + 1)
2
Which shows the next sum equals the functionat the next natural number
![Page 61: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/61.jpg)
Prove∑
k = n(n−1)2
Induction: Pretend
0 + 1 + 2 + · · ·+ (n − 2) + (n − 1) =n(n − 1)
2
for some n ≥ 0Then the next sum is:
[0 + 1 + · · ·+ (n − 1)] + n =n(n − 1)
2+ n
=n(n − 1)
2+
2n2
=n(n + 1)
2
Which shows the next sum equals the functionat the next natural number
![Page 62: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/62.jpg)
Prove∑
k = n(n−1)2
Induction: Pretend
0 + 1 + 2 + · · ·+ (n − 2) + (n − 1) =n(n − 1)
2
for some n ≥ 0
Then the next sum is:
[0 + 1 + · · ·+ (n − 1)] + n =n(n − 1)
2+ n
=n(n − 1)
2+
2n2
=n(n + 1)
2
Which shows the next sum equals the functionat the next natural number
![Page 63: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/63.jpg)
Prove∑
k = n(n−1)2
Induction: Pretend
0 + 1 + 2 + · · ·+ (n − 2) + (n − 1) =n(n − 1)
2
for some n ≥ 0Then the next sum is:
[0 + 1 + · · ·+ (n − 1)] + n =n(n − 1)
2+ n
=n(n − 1)
2+
2n2
=n(n + 1)
2
Which shows the next sum equals the functionat the next natural number
![Page 64: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/64.jpg)
Prove∑
k = n(n−1)2
Induction: Pretend
0 + 1 + 2 + · · ·+ (n − 2) + (n − 1) =n(n − 1)
2
for some n ≥ 0Then the next sum is:
[0 + 1 + · · ·+ (n − 1)] + n =n(n − 1)
2+ n
=n(n − 1)
2+
2n2
=n(n + 1)
2
Which shows the next sum equals the functionat the next natural number
![Page 65: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/65.jpg)
Prove∑
k = n(n−1)2
Induction: Pretend
0 + 1 + 2 + · · ·+ (n − 2) + (n − 1) =n(n − 1)
2
for some n ≥ 0Then the next sum is:
[0 + 1 + · · ·+ (n − 1)] + n
=n(n − 1)
2+ n
=n(n − 1)
2+
2n2
=n(n + 1)
2
Which shows the next sum equals the functionat the next natural number
![Page 66: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/66.jpg)
Prove∑
k = n(n−1)2
Induction: Pretend
0 + 1 + 2 + · · ·+ (n − 2) + (n − 1) =n(n − 1)
2
for some n ≥ 0Then the next sum is:
[0 + 1 + · · ·+ (n − 1)] + n =n(n − 1)
2+ n
=n(n − 1)
2+
2n2
=n(n + 1)
2
Which shows the next sum equals the functionat the next natural number
![Page 67: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/67.jpg)
Prove∑
k = n(n−1)2
Induction: Pretend
0 + 1 + 2 + · · ·+ (n − 2) + (n − 1) =n(n − 1)
2
for some n ≥ 0Then the next sum is:
[0 + 1 + · · ·+ (n − 1)] + n =n(n − 1)
2+ n
=n(n − 1)
2+
2n2
=n(n + 1)
2
Which shows the next sum equals the functionat the next natural number
![Page 68: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/68.jpg)
Prove∑
k = n(n−1)2
Induction: Pretend
0 + 1 + 2 + · · ·+ (n − 2) + (n − 1) =n(n − 1)
2
for some n ≥ 0Then the next sum is:
[0 + 1 + · · ·+ (n − 1)] + n =n(n − 1)
2+ n
=n(n − 1)
2+
2n2
=n(n + 1)
2
Which shows the next sum equals the functionat the next natural number
![Page 69: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/69.jpg)
Inductive Template For Functional Equality
To prove “f (n) = g(n)” is true for all natural numbers n
Basis: Show: both functions map 0 to the same value, thatis,
f (0) = g(0)
Induction: Show: if f (n) = g(n) for some n ≥ 0, thenf (n + 1) = g(n + 1)
![Page 70: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/70.jpg)
Inductive Template For Functional Equality
To prove “f (n) = g(n)” is true for all natural numbers n
Basis: Show: both functions map 0 to the same value, thatis,
f (0) = g(0)
Induction: Show: if f (n) = g(n) for some n ≥ 0, thenf (n + 1) = g(n + 1)
![Page 71: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/71.jpg)
Inductive Template For Functional Equality
To prove “f (n) = g(n)” is true for all natural numbers n
Basis: Show: both functions map 0 to the same value, thatis,
f (0) = g(0)
Induction: Show: if f (n) = g(n) for some n ≥ 0, thenf (n + 1) = g(n + 1)
![Page 72: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/72.jpg)
Inductive Template For Functional Equality
To prove “f (n) = g(n)” is true for all natural numbers n
Basis: Show: both functions map 0 to the same value, thatis,
f (0) = g(0)
Induction: Show: if f (n) = g(n) for some n ≥ 0, thenf (n + 1) = g(n + 1)
![Page 73: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/73.jpg)
Quotation
The so-called law of induction cannot possibly be a lawof logic, since it is obviously a proposition with a sense.Nor, therefore, can it be an a priori law.
Ludwig Wittenstein, "Tractatus Logico-Philosophicus"Austrian-British philosoper (1889 - 1951)
![Page 74: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/74.jpg)
Checking Solutions to Algebraic Equations
Pretend you want to check that x = 4 is a solution to thelinear equation
3x − 5 = 7
You can do this by substitiuting 4 for x on the left-hand sideand show the result is the right-hand side
3(4)− 5 = 12− 5 = 7
![Page 75: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/75.jpg)
Checking Solutions to Algebraic Equations
Pretend you want to check that x = 4 is a solution to thelinear equation
3x − 5 = 7
You can do this by substitiuting 4 for x on the left-hand sideand show the result is the right-hand side
3(4)− 5 = 12− 5 = 7
![Page 76: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/76.jpg)
Checking Solutions to Algebraic Equations
Pretend you want to check that x = 4 is a solution to thelinear equation
3x − 5 = 7
You can do this by substitiuting 4 for x on the left-hand sideand show the result is the right-hand side
3(4)− 5 = 12− 5 = 7
![Page 77: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/77.jpg)
Checking Solutions to Algebraic Equations
Pretend you want to check that x = (1 +√5)/2 is a solution
to the quadratic equation
x2 = x + 1
You can do this by substitiuting (1 +√5)/2 for x in both the
left-hand and right-hand sides and show the results are equal
The left-hand side is(1 +√5
2
)2
=1 + 2
√5 + 5
4=
3 +√5
2
The right-hand side is
1 +√5
2+ 1 =
1 +√5
2+
22
=3 +√5
2
![Page 78: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/78.jpg)
Checking Solutions to Algebraic Equations
Pretend you want to check that x = (1 +√5)/2 is a solution
to the quadratic equation
x2 = x + 1
You can do this by substitiuting (1 +√5)/2 for x in both the
left-hand and right-hand sides and show the results are equal
The left-hand side is(1 +√5
2
)2
=1 + 2
√5 + 5
4=
3 +√5
2
The right-hand side is
1 +√5
2+ 1 =
1 +√5
2+
22
=3 +√5
2
![Page 79: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/79.jpg)
Checking Solutions to Algebraic Equations
Pretend you want to check that x = (1 +√5)/2 is a solution
to the quadratic equation
x2 = x + 1
You can do this by substitiuting (1 +√5)/2 for x in both the
left-hand and right-hand sides and show the results are equal
The left-hand side is(1 +√5
2
)2
=1 + 2
√5 + 5
4=
3 +√5
2
The right-hand side is
1 +√5
2+ 1 =
1 +√5
2+
22
=3 +√5
2
![Page 80: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/80.jpg)
Checking Solutions to Algebraic Equations
Pretend you want to check that x = (1 +√5)/2 is a solution
to the quadratic equation
x2 = x + 1
You can do this by substitiuting (1 +√5)/2 for x in both the
left-hand and right-hand sides and show the results are equalThe left-hand side is(
1 +√5
2
)2
=1 + 2
√5 + 5
4=
3 +√5
2
The right-hand side is
1 +√5
2+ 1 =
1 +√5
2+
22
=3 +√5
2
![Page 81: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/81.jpg)
Checking Solutions to Algebraic Equations
Pretend you want to check that x = (1 +√5)/2 is a solution
to the quadratic equation
x2 = x + 1
You can do this by substitiuting (1 +√5)/2 for x in both the
left-hand and right-hand sides and show the results are equalThe left-hand side is(
1 +√5
2
)2
=1 + 2
√5 + 5
4=
3 +√5
2
The right-hand side is
1 +√5
2+ 1 =
1 +√5
2+
22
=3 +√5
2
![Page 82: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/82.jpg)
Checking Solutions to Recurrence Equations
The same basic idea can be used to check a proposed solutionto a recurrence equationSuppose you want to show that mn = 2n − 1 is a solution tothe recurrence
mn = 2mn−1 + 1
You can do this by substitiuting 2n − 1 for mn in the left-handside and 2n−1 − 1 for mn−1 in the right-hand sides and showthe results are equal
The left-hand side is2n − 1
The right-hand side is
2(2n−1 − 1
)+ 1 = (2n − 2) + 1 = 2n − 1
![Page 83: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/83.jpg)
Checking Solutions to Recurrence Equations
The same basic idea can be used to check a proposed solutionto a recurrence equation
Suppose you want to show that mn = 2n − 1 is a solution tothe recurrence
mn = 2mn−1 + 1
You can do this by substitiuting 2n − 1 for mn in the left-handside and 2n−1 − 1 for mn−1 in the right-hand sides and showthe results are equal
The left-hand side is2n − 1
The right-hand side is
2(2n−1 − 1
)+ 1 = (2n − 2) + 1 = 2n − 1
![Page 84: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/84.jpg)
Checking Solutions to Recurrence Equations
The same basic idea can be used to check a proposed solutionto a recurrence equationSuppose you want to show that mn = 2n − 1 is a solution tothe recurrence
mn = 2mn−1 + 1
You can do this by substitiuting 2n − 1 for mn in the left-handside and 2n−1 − 1 for mn−1 in the right-hand sides and showthe results are equal
The left-hand side is2n − 1
The right-hand side is
2(2n−1 − 1
)+ 1 = (2n − 2) + 1 = 2n − 1
![Page 85: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/85.jpg)
Checking Solutions to Recurrence Equations
The same basic idea can be used to check a proposed solutionto a recurrence equationSuppose you want to show that mn = 2n − 1 is a solution tothe recurrence
mn = 2mn−1 + 1
You can do this by substitiuting 2n − 1 for mn in the left-handside and 2n−1 − 1 for mn−1 in the right-hand sides and showthe results are equal
The left-hand side is2n − 1
The right-hand side is
2(2n−1 − 1
)+ 1 = (2n − 2) + 1 = 2n − 1
![Page 86: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/86.jpg)
Checking Solutions to Recurrence Equations
The same basic idea can be used to check a proposed solutionto a recurrence equationSuppose you want to show that mn = 2n − 1 is a solution tothe recurrence
mn = 2mn−1 + 1
You can do this by substitiuting 2n − 1 for mn in the left-handside and 2n−1 − 1 for mn−1 in the right-hand sides and showthe results are equal
The left-hand side is2n − 1
The right-hand side is
2(2n−1 − 1
)+ 1 = (2n − 2) + 1 = 2n − 1
![Page 87: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/87.jpg)
Checking Solutions to Recurrence Equations
The same basic idea can be used to check a proposed solutionto a recurrence equationSuppose you want to show that mn = 2n − 1 is a solution tothe recurrence
mn = 2mn−1 + 1
You can do this by substitiuting 2n − 1 for mn in the left-handside and 2n−1 − 1 for mn−1 in the right-hand sides and showthe results are equal
The left-hand side is2n − 1
The right-hand side is
2(2n−1 − 1
)+ 1 = (2n − 2) + 1 = 2n − 1
![Page 88: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/88.jpg)
Checking Solutions to Recurrence Equations
There’s is one problem with thisFor any constant c , you can show mn = c2n − 1 is a solutionto the recurrence
mn = 2mn−1 + 1
The left-hand side would be
c2n − 1
And the right-hand side would be
2(c2n−1 − 1
)+ 1 = c2n − 1
An initial value nails down the solution: If m0 = 0, thenc20 − 1 = c − 1 = 0 forces c to be 1
![Page 89: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/89.jpg)
Checking Solutions to Recurrence Equations
There’s is one problem with this
For any constant c , you can show mn = c2n − 1 is a solutionto the recurrence
mn = 2mn−1 + 1
The left-hand side would be
c2n − 1
And the right-hand side would be
2(c2n−1 − 1
)+ 1 = c2n − 1
An initial value nails down the solution: If m0 = 0, thenc20 − 1 = c − 1 = 0 forces c to be 1
![Page 90: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/90.jpg)
Checking Solutions to Recurrence Equations
There’s is one problem with thisFor any constant c , you can show mn = c2n − 1 is a solutionto the recurrence
mn = 2mn−1 + 1
The left-hand side would be
c2n − 1
And the right-hand side would be
2(c2n−1 − 1
)+ 1 = c2n − 1
An initial value nails down the solution: If m0 = 0, thenc20 − 1 = c − 1 = 0 forces c to be 1
![Page 91: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/91.jpg)
Checking Solutions to Recurrence Equations
There’s is one problem with thisFor any constant c , you can show mn = c2n − 1 is a solutionto the recurrence
mn = 2mn−1 + 1
The left-hand side would be
c2n − 1
And the right-hand side would be
2(c2n−1 − 1
)+ 1 = c2n − 1
An initial value nails down the solution: If m0 = 0, thenc20 − 1 = c − 1 = 0 forces c to be 1
![Page 92: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/92.jpg)
Checking Solutions to Recurrence Equations
There’s is one problem with thisFor any constant c , you can show mn = c2n − 1 is a solutionto the recurrence
mn = 2mn−1 + 1
The left-hand side would be
c2n − 1
And the right-hand side would be
2(c2n−1 − 1
)+ 1 = c2n − 1
An initial value nails down the solution: If m0 = 0, thenc20 − 1 = c − 1 = 0 forces c to be 1
![Page 93: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/93.jpg)
Checking Solutions to Recurrence Equations
There’s is one problem with thisFor any constant c , you can show mn = c2n − 1 is a solutionto the recurrence
mn = 2mn−1 + 1
The left-hand side would be
c2n − 1
And the right-hand side would be
2(c2n−1 − 1
)+ 1 = c2n − 1
An initial value nails down the solution: If m0 = 0, thenc20 − 1 = c − 1 = 0 forces c to be 1
![Page 94: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/94.jpg)
Induction on The Triangular Sequence
Consider the recurrence for the triangular numbers
tn = tn−1 + (n − 1)
To show that tn = n(n − 1)/2 solves the recurrence substituteinto the left-hand and right hand sides
The left-hand side isn(n − 1)/2
The right-hand side is
(n − 1)(n − 2)2
+ (n − 1) =(n − 1)(n − 2)
2+
2(n − 1)2
=(n − 1)(n − 2) + 2(n − 1)
2
=n(n − 1)
2
To complete the induction, establish the basis:t0 = 0(0− 1)/2 = 0
![Page 95: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/95.jpg)
Induction on The Triangular Sequence
Consider the recurrence for the triangular numbers
tn = tn−1 + (n − 1)
To show that tn = n(n − 1)/2 solves the recurrence substituteinto the left-hand and right hand sides
The left-hand side isn(n − 1)/2
The right-hand side is
(n − 1)(n − 2)2
+ (n − 1) =(n − 1)(n − 2)
2+
2(n − 1)2
=(n − 1)(n − 2) + 2(n − 1)
2
=n(n − 1)
2
To complete the induction, establish the basis:t0 = 0(0− 1)/2 = 0
![Page 96: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/96.jpg)
Induction on The Triangular Sequence
Consider the recurrence for the triangular numbers
tn = tn−1 + (n − 1)
To show that tn = n(n − 1)/2 solves the recurrence substituteinto the left-hand and right hand sides
The left-hand side isn(n − 1)/2
The right-hand side is
(n − 1)(n − 2)2
+ (n − 1) =(n − 1)(n − 2)
2+
2(n − 1)2
=(n − 1)(n − 2) + 2(n − 1)
2
=n(n − 1)
2
To complete the induction, establish the basis:t0 = 0(0− 1)/2 = 0
![Page 97: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/97.jpg)
Induction on The Triangular Sequence
Consider the recurrence for the triangular numbers
tn = tn−1 + (n − 1)
To show that tn = n(n − 1)/2 solves the recurrence substituteinto the left-hand and right hand sides
The left-hand side isn(n − 1)/2
The right-hand side is
(n − 1)(n − 2)2
+ (n − 1) =(n − 1)(n − 2)
2+
2(n − 1)2
=(n − 1)(n − 2) + 2(n − 1)
2
=n(n − 1)
2
To complete the induction, establish the basis:t0 = 0(0− 1)/2 = 0
![Page 98: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/98.jpg)
Induction on The Triangular Sequence
Consider the recurrence for the triangular numbers
tn = tn−1 + (n − 1)
To show that tn = n(n − 1)/2 solves the recurrence substituteinto the left-hand and right hand sides
The left-hand side isn(n − 1)/2
The right-hand side is
(n − 1)(n − 2)2
+ (n − 1) =(n − 1)(n − 2)
2+
2(n − 1)2
=(n − 1)(n − 2) + 2(n − 1)
2
=n(n − 1)
2
To complete the induction, establish the basis:t0 = 0(0− 1)/2 = 0
![Page 99: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/99.jpg)
Induction on The Triangular Sequence
Consider the recurrence for the triangular numbers
tn = tn−1 + (n − 1)
To show that tn = n(n − 1)/2 solves the recurrence substituteinto the left-hand and right hand sides
The left-hand side isn(n − 1)/2
The right-hand side is
(n − 1)(n − 2)2
+ (n − 1) =(n − 1)(n − 2)
2+
2(n − 1)2
=(n − 1)(n − 2) + 2(n − 1)
2
=n(n − 1)
2
To complete the induction, establish the basis:t0 = 0(0− 1)/2 = 0
![Page 100: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/100.jpg)
Induction on The Mersenne Sequence
Consider the recurrence for the Mersenne numbers
mn = 2mn−1 + 1
or, relabeling subscripts
mn+1 = 2mn + 1
If mn = 2n − 1, then
mn+1 = 2mn + 1 = 2(2n − 1) + 1
Which can be rewritten as
mn+1 = 2n+1 − 1
The function 2n − 1 maps 0 to 0 which matches the initialMersenne number m0 = 0Since both the basis and the inductive conditional are bothtrue, the Mersenne numbers mn can be computed by thefunction m(n) = 2n − 1 for all natural numbers n
![Page 101: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/101.jpg)
Induction on The Mersenne Sequence
Consider the recurrence for the Mersenne numbers
mn = 2mn−1 + 1
or, relabeling subscripts
mn+1 = 2mn + 1
If mn = 2n − 1, then
mn+1 = 2mn + 1 = 2(2n − 1) + 1
Which can be rewritten as
mn+1 = 2n+1 − 1
The function 2n − 1 maps 0 to 0 which matches the initialMersenne number m0 = 0Since both the basis and the inductive conditional are bothtrue, the Mersenne numbers mn can be computed by thefunction m(n) = 2n − 1 for all natural numbers n
![Page 102: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/102.jpg)
Induction on The Mersenne Sequence
Consider the recurrence for the Mersenne numbers
mn = 2mn−1 + 1
or, relabeling subscripts
mn+1 = 2mn + 1
If mn = 2n − 1, then
mn+1 = 2mn + 1 = 2(2n − 1) + 1
Which can be rewritten as
mn+1 = 2n+1 − 1
The function 2n − 1 maps 0 to 0 which matches the initialMersenne number m0 = 0Since both the basis and the inductive conditional are bothtrue, the Mersenne numbers mn can be computed by thefunction m(n) = 2n − 1 for all natural numbers n
![Page 103: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/103.jpg)
Induction on The Mersenne Sequence
Consider the recurrence for the Mersenne numbers
mn = 2mn−1 + 1
or, relabeling subscripts
mn+1 = 2mn + 1
If mn = 2n − 1, then
mn+1 = 2mn + 1 = 2(2n − 1) + 1
Which can be rewritten as
mn+1 = 2n+1 − 1
The function 2n − 1 maps 0 to 0 which matches the initialMersenne number m0 = 0Since both the basis and the inductive conditional are bothtrue, the Mersenne numbers mn can be computed by thefunction m(n) = 2n − 1 for all natural numbers n
![Page 104: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/104.jpg)
Induction on The Mersenne Sequence
Consider the recurrence for the Mersenne numbers
mn = 2mn−1 + 1
or, relabeling subscripts
mn+1 = 2mn + 1
If mn = 2n − 1, then
mn+1 = 2mn + 1 = 2(2n − 1) + 1
Which can be rewritten as
mn+1 = 2n+1 − 1
The function 2n − 1 maps 0 to 0 which matches the initialMersenne number m0 = 0
Since both the basis and the inductive conditional are bothtrue, the Mersenne numbers mn can be computed by thefunction m(n) = 2n − 1 for all natural numbers n
![Page 105: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/105.jpg)
Induction on The Mersenne Sequence
Consider the recurrence for the Mersenne numbers
mn = 2mn−1 + 1
or, relabeling subscripts
mn+1 = 2mn + 1
If mn = 2n − 1, then
mn+1 = 2mn + 1 = 2(2n − 1) + 1
Which can be rewritten as
mn+1 = 2n+1 − 1
The function 2n − 1 maps 0 to 0 which matches the initialMersenne number m0 = 0Since both the basis and the inductive conditional are bothtrue, the Mersenne numbers mn can be computed by thefunction m(n) = 2n − 1 for all natural numbers n
![Page 106: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/106.jpg)
Induction on The Fermat Sequence
Consider the recurrence for the Fermat numbers
rn = (rn−1 − 1)2 + 1
or, relabeling subscripts
rn+1 = (rn − 1)2 + 1
If rn = 22n+ 1, then
rn+1 = ((22n+ 1)− 1)2 + 1
Which can be rewritten as
rn+1 = 22n+1+ 1
The function 22n+ 1 maps 0 to 3 which matches the initial
Fermat number r0 = 3Since both the basis and the inductive conditional are bothtrue, the Fermat numbers rn can be computed by the functionr(n) = 22n
+ 1 for all natural numbers n
![Page 107: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/107.jpg)
Induction on The Fermat Sequence
Consider the recurrence for the Fermat numbers
rn = (rn−1 − 1)2 + 1
or, relabeling subscripts
rn+1 = (rn − 1)2 + 1
If rn = 22n+ 1, then
rn+1 = ((22n+ 1)− 1)2 + 1
Which can be rewritten as
rn+1 = 22n+1+ 1
The function 22n+ 1 maps 0 to 3 which matches the initial
Fermat number r0 = 3Since both the basis and the inductive conditional are bothtrue, the Fermat numbers rn can be computed by the functionr(n) = 22n
+ 1 for all natural numbers n
![Page 108: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/108.jpg)
Induction on The Fermat Sequence
Consider the recurrence for the Fermat numbers
rn = (rn−1 − 1)2 + 1
or, relabeling subscripts
rn+1 = (rn − 1)2 + 1
If rn = 22n+ 1, then
rn+1 = ((22n+ 1)− 1)2 + 1
Which can be rewritten as
rn+1 = 22n+1+ 1
The function 22n+ 1 maps 0 to 3 which matches the initial
Fermat number r0 = 3Since both the basis and the inductive conditional are bothtrue, the Fermat numbers rn can be computed by the functionr(n) = 22n
+ 1 for all natural numbers n
![Page 109: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/109.jpg)
Induction on The Fermat Sequence
Consider the recurrence for the Fermat numbers
rn = (rn−1 − 1)2 + 1
or, relabeling subscripts
rn+1 = (rn − 1)2 + 1
If rn = 22n+ 1, then
rn+1 = ((22n+ 1)− 1)2 + 1
Which can be rewritten as
rn+1 = 22n+1+ 1
The function 22n+ 1 maps 0 to 3 which matches the initial
Fermat number r0 = 3Since both the basis and the inductive conditional are bothtrue, the Fermat numbers rn can be computed by the functionr(n) = 22n
+ 1 for all natural numbers n
![Page 110: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/110.jpg)
Induction on The Fermat Sequence
Consider the recurrence for the Fermat numbers
rn = (rn−1 − 1)2 + 1
or, relabeling subscripts
rn+1 = (rn − 1)2 + 1
If rn = 22n+ 1, then
rn+1 = ((22n+ 1)− 1)2 + 1
Which can be rewritten as
rn+1 = 22n+1+ 1
The function 22n+ 1 maps 0 to 3 which matches the initial
Fermat number r0 = 3
Since both the basis and the inductive conditional are bothtrue, the Fermat numbers rn can be computed by the functionr(n) = 22n
+ 1 for all natural numbers n
![Page 111: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/111.jpg)
Induction on The Fermat Sequence
Consider the recurrence for the Fermat numbers
rn = (rn−1 − 1)2 + 1
or, relabeling subscripts
rn+1 = (rn − 1)2 + 1
If rn = 22n+ 1, then
rn+1 = ((22n+ 1)− 1)2 + 1
Which can be rewritten as
rn+1 = 22n+1+ 1
The function 22n+ 1 maps 0 to 3 which matches the initial
Fermat number r0 = 3Since both the basis and the inductive conditional are bothtrue, the Fermat numbers rn can be computed by the functionr(n) = 22n
+ 1 for all natural numbers n
![Page 112: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/112.jpg)
Quotation
MacPherson told me that my theorem can be viewedas blah blah blah Grothendick blah blah blah, whichmakes it much more respectable. I think someintuition leaks out in every step of an induction proof.
Jim Propp, American mathematician
![Page 113: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/113.jpg)
Induction on Sums
To prove a summation∑
ak can be expressed as a functiong(·) on the natural numbers,
Establish a basis for induction, that is, show0∑
k=0
ak = a0 = g(0)
Prove the inductive conditional: If
Ifn−1∑k=0
ak = g(n),
then
n∑k=0
ak =
[n−1∑k=0
ak
]+ an
= g(n) + an
= g(n + 1)
![Page 114: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/114.jpg)
Induction on Sums
To prove a summation∑
ak can be expressed as a functiong(·) on the natural numbers,
Establish a basis for induction, that is, show0∑
k=0
ak = a0 = g(0)
Prove the inductive conditional: If
Ifn−1∑k=0
ak = g(n),
then
n∑k=0
ak =
[n−1∑k=0
ak
]+ an
= g(n) + an
= g(n + 1)
![Page 115: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/115.jpg)
Induction on Sums
To prove a summation∑
ak can be expressed as a functiong(·) on the natural numbers,
Establish a basis for induction, that is, show0∑
k=0
ak = a0 = g(0)
Prove the inductive conditional: If
Ifn−1∑k=0
ak = g(n),
then
n∑k=0
ak =
[n−1∑k=0
ak
]+ an
= g(n) + an
= g(n + 1)
![Page 116: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/116.jpg)
Induction on Sums
To prove a summation∑
ak can be expressed as a functiong(·) on the natural numbers,
Establish a basis for induction, that is, show0∑
k=0
ak = a0 = g(0)
Prove the inductive conditional: If
Ifn−1∑k=0
ak = g(n),
then
n∑k=0
ak =
[n−1∑k=0
ak
]+ an
= g(n) + an
= g(n + 1)
![Page 117: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/117.jpg)
Induction on Sums
To prove a summation∑
ak can be expressed as a functiong(·) on the natural numbers,
Establish a basis for induction, that is, show0∑
k=0
ak = a0 = g(0)
Prove the inductive conditional: If
Ifn−1∑k=0
ak = g(n),
then
n∑k=0
ak =
[n−1∑k=0
ak
]+ an
= g(n) + an
= g(n + 1)
![Page 118: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/118.jpg)
Induction on Sums
To prove a summation∑
ak can be expressed as a functiong(·) on the natural numbers,
Establish a basis for induction, that is, show0∑
k=0
ak = a0 = g(0)
Prove the inductive conditional: If
Ifn−1∑k=0
ak = g(n),
thenn∑
k=0
ak
=
[n−1∑k=0
ak
]+ an
= g(n) + an
= g(n + 1)
![Page 119: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/119.jpg)
Induction on Sums
To prove a summation∑
ak can be expressed as a functiong(·) on the natural numbers,
Establish a basis for induction, that is, show0∑
k=0
ak = a0 = g(0)
Prove the inductive conditional: If
Ifn−1∑k=0
ak = g(n),
thenn∑
k=0
ak =
[n−1∑k=0
ak
]+ an
= g(n) + an
= g(n + 1)
![Page 120: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/120.jpg)
Induction on Sums
To prove a summation∑
ak can be expressed as a functiong(·) on the natural numbers,
Establish a basis for induction, that is, show0∑
k=0
ak = a0 = g(0)
Prove the inductive conditional: If
Ifn−1∑k=0
ak = g(n),
thenn∑
k=0
ak =
[n−1∑k=0
ak
]+ an
= g(n) + an
= g(n + 1)
![Page 121: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/121.jpg)
Induction on Sums
To prove a summation∑
ak can be expressed as a functiong(·) on the natural numbers,
Establish a basis for induction, that is, show0∑
k=0
ak = a0 = g(0)
Prove the inductive conditional: If
Ifn−1∑k=0
ak = g(n),
thenn∑
k=0
ak =
[n−1∑k=0
ak
]+ an
= g(n) + an
= g(n + 1)
![Page 122: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/122.jpg)
Alice Sum Math Induction Proof
The Alice sum formula
n−1∑k=0
1 = n
can be established by mathematical induction
Basis: For n = 0
The sumn−1∑k=0
1 =0−1∑k=0
1
is empty and equal to 0The value on the right hand, n, is 0 too
![Page 123: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/123.jpg)
Alice Sum Math Induction Proof
The Alice sum formula
n−1∑k=0
1 = n
can be established by mathematical induction
Basis: For n = 0
The sumn−1∑k=0
1 =0−1∑k=0
1
is empty and equal to 0The value on the right hand, n, is 0 too
![Page 124: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/124.jpg)
Alice Sum Math Induction Proof
The Alice sum formula
n−1∑k=0
1 = n
can be established by mathematical induction
Basis: For n = 0
The sumn−1∑k=0
1 =0−1∑k=0
1
is empty and equal to 0The value on the right hand, n, is 0 too
![Page 125: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/125.jpg)
Alice Sum Math Induction Proof
The Alice sum formula
n−1∑k=0
1 = n
can be established by mathematical induction
Basis: For n = 0The sum
n−1∑k=0
1 =0−1∑k=0
1
is empty and equal to 0
The value on the right hand, n, is 0 too
![Page 126: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/126.jpg)
Alice Sum Math Induction Proof
The Alice sum formula
n−1∑k=0
1 = n
can be established by mathematical induction
Basis: For n = 0The sum
n−1∑k=0
1 =0−1∑k=0
1
is empty and equal to 0The value on the right hand, n, is 0 too
![Page 127: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/127.jpg)
Alice Sum Math Induction Proof
Induction: Show that
Ifn−1∑k=0
1 = n,
thenn∑
k=0
1 = n + 1
By computing
n∑k=0
1 =
[n−1∑k=0
1
]+ 1
= n + 1
![Page 128: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/128.jpg)
Alice Sum Math Induction Proof
Induction: Show that
Ifn−1∑k=0
1 = n,
thenn∑
k=0
1 = n + 1
By computing
n∑k=0
1 =
[n−1∑k=0
1
]+ 1
= n + 1
![Page 129: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/129.jpg)
Alice Sum Math Induction Proof
Induction: Show thatIf
n−1∑k=0
1 = n,
thenn∑
k=0
1 = n + 1
By computing
n∑k=0
1 =
[n−1∑k=0
1
]+ 1
= n + 1
![Page 130: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/130.jpg)
Alice Sum Math Induction Proof
Induction: Show thatIf
n−1∑k=0
1 = n,
thenn∑
k=0
1 = n + 1
By computing
n∑k=0
1 =
[n−1∑k=0
1
]+ 1
= n + 1
![Page 131: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/131.jpg)
Alice Sum Math Induction Proof
Induction: Show thatIf
n−1∑k=0
1 = n,
thenn∑
k=0
1 = n + 1
By computing
n∑k=0
1
=
[n−1∑k=0
1
]+ 1
= n + 1
![Page 132: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/132.jpg)
Alice Sum Math Induction Proof
Induction: Show thatIf
n−1∑k=0
1 = n,
thenn∑
k=0
1 = n + 1
By computing
n∑k=0
1 =
[n−1∑k=0
1
]+ 1
= n + 1
![Page 133: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/133.jpg)
Alice Sum Math Induction Proof
Induction: Show thatIf
n−1∑k=0
1 = n,
thenn∑
k=0
1 = n + 1
By computing
n∑k=0
1 =
[n−1∑k=0
1
]+ 1
= n + 1
![Page 134: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/134.jpg)
Gauss Sum Math Induction Proof
The Gauss sum formula
n−1∑k=0
k = 0 + 1 + 2 + 3 + · · ·+ (n − 1) =n(n − 1)
2
can be established by mathematical induction
Basis: For n = 0 the sum on the left is empty, and so equalto 0. And the right hand side is 0(0− 1)/2 = 0 too
Induction: If∑n−1
k=0 k = n(n − 1)/2, then
n∑k=0
k =
[n−1∑k=0
k
]+ n = (n + 1)n/2
![Page 135: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/135.jpg)
Gauss Sum Math Induction Proof
The Gauss sum formula
n−1∑k=0
k = 0 + 1 + 2 + 3 + · · ·+ (n − 1) =n(n − 1)
2
can be established by mathematical induction
Basis: For n = 0 the sum on the left is empty, and so equalto 0. And the right hand side is 0(0− 1)/2 = 0 too
Induction: If∑n−1
k=0 k = n(n − 1)/2, then
n∑k=0
k =
[n−1∑k=0
k
]+ n = (n + 1)n/2
![Page 136: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/136.jpg)
Gauss Sum Math Induction Proof
The Gauss sum formula
n−1∑k=0
k = 0 + 1 + 2 + 3 + · · ·+ (n − 1) =n(n − 1)
2
can be established by mathematical induction
Basis: For n = 0 the sum on the left is empty, and so equalto 0. And the right hand side is 0(0− 1)/2 = 0 too
Induction: If∑n−1
k=0 k = n(n − 1)/2, then
n∑k=0
k =
[n−1∑k=0
k
]+ n = (n + 1)n/2
![Page 137: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/137.jpg)
Gauss Sum Math Induction Proof
The Gauss sum formula
n−1∑k=0
k = 0 + 1 + 2 + 3 + · · ·+ (n − 1) =n(n − 1)
2
can be established by mathematical induction
Basis: For n = 0 the sum on the left is empty, and so equalto 0. And the right hand side is 0(0− 1)/2 = 0 too
Induction: If∑n−1
k=0 k = n(n − 1)/2, then
n∑k=0
k =
[n−1∑k=0
k
]+ n = (n + 1)n/2
![Page 138: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/138.jpg)
Zeno Sum Math Induction Proof
The Zeno sum formula
n−1∑k=0
2k = 1 + 2 + 4 + · · ·+ 2n−1 = 2n − 1
can be established by mathematical induction
Basis: For n = 0 the sum on the left is empty, and so equalto 0. And the right hand side n is 0 too
Induction: If∑n−1
k=0 2k = 2n − 1, then
n∑k=0
2k =
[n−1∑k=0
2k
]+ 2n = 2n+1 − 1
![Page 139: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/139.jpg)
Zeno Sum Math Induction Proof
The Zeno sum formula
n−1∑k=0
2k = 1 + 2 + 4 + · · ·+ 2n−1 = 2n − 1
can be established by mathematical induction
Basis: For n = 0 the sum on the left is empty, and so equalto 0. And the right hand side n is 0 too
Induction: If∑n−1
k=0 2k = 2n − 1, then
n∑k=0
2k =
[n−1∑k=0
2k
]+ 2n = 2n+1 − 1
![Page 140: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/140.jpg)
Zeno Sum Math Induction Proof
The Zeno sum formula
n−1∑k=0
2k = 1 + 2 + 4 + · · ·+ 2n−1 = 2n − 1
can be established by mathematical induction
Basis: For n = 0 the sum on the left is empty, and so equalto 0. And the right hand side n is 0 too
Induction: If∑n−1
k=0 2k = 2n − 1, then
n∑k=0
2k =
[n−1∑k=0
2k
]+ 2n = 2n+1 − 1
![Page 141: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/141.jpg)
Zeno Sum Math Induction Proof
The Zeno sum formula
n−1∑k=0
2k = 1 + 2 + 4 + · · ·+ 2n−1 = 2n − 1
can be established by mathematical induction
Basis: For n = 0 the sum on the left is empty, and so equalto 0. And the right hand side n is 0 too
Induction: If∑n−1
k=0 2k = 2n − 1, then
n∑k=0
2k =
[n−1∑k=0
2k
]+ 2n = 2n+1 − 1
![Page 142: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/142.jpg)
Sum of the Odd Integers is a Square
The ancient formulan∑
k=1
(2k − 1) = 1 + 3 + 5 + · · ·+ (2n − 1) = n2
can be established by mathematical induction
Basis: For n = 0 the sum on the left is empty, and so equalto 0. And the right hand side n is 0 too.
![Page 143: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/143.jpg)
Sum of the Odd Integers is a Square
The ancient formulan∑
k=1
(2k − 1) = 1 + 3 + 5 + · · ·+ (2n − 1) = n2
can be established by mathematical induction
Basis: For n = 0 the sum on the left is empty, and so equalto 0. And the right hand side n is 0 too.
![Page 144: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/144.jpg)
Sum of the Odd Integers is a Square
The ancient formulan∑
k=1
(2k − 1) = 1 + 3 + 5 + · · ·+ (2n − 1) = n2
can be established by mathematical induction
Basis: For n = 0 the sum on the left is empty, and so equalto 0. And the right hand side n is 0 too.
![Page 145: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/145.jpg)
Sum of the Odd Integers is a Square
n∑k=1
(2k − 1) = 1 + 3 + 5 + · · ·+ (2n − 1) = n2
Induction: If∑n
k=1(2k − 1) = n2, then
n+1∑k=1
(2k − 1) =
[n∑
k=1
(2k − 1)
]+ 2(n + 1)− 1
= n2 + 2(n + 1)− 1
= n2 + 2n + 1
= (n + 1)2
![Page 146: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/146.jpg)
Sum of the Odd Integers is a Square
n∑k=1
(2k − 1) = 1 + 3 + 5 + · · ·+ (2n − 1) = n2
Induction: If∑n
k=1(2k − 1) = n2, then
n+1∑k=1
(2k − 1) =
[n∑
k=1
(2k − 1)
]+ 2(n + 1)− 1
= n2 + 2(n + 1)− 1
= n2 + 2n + 1
= (n + 1)2
![Page 147: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/147.jpg)
Sum of the Odd Integers is a Square
n∑k=1
(2k − 1) = 1 + 3 + 5 + · · ·+ (2n − 1) = n2
Induction: If∑n
k=1(2k − 1) = n2, then
n+1∑k=1
(2k − 1) =
[n∑
k=1
(2k − 1)
]+ 2(n + 1)− 1
= n2 + 2(n + 1)− 1
= n2 + 2n + 1
= (n + 1)2
![Page 148: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/148.jpg)
Sum of the Odd Integers is a Square
n∑k=1
(2k − 1) = 1 + 3 + 5 + · · ·+ (2n − 1) = n2
Induction: If∑n
k=1(2k − 1) = n2, then
n+1∑k=1
(2k − 1)
=
[n∑
k=1
(2k − 1)
]+ 2(n + 1)− 1
= n2 + 2(n + 1)− 1
= n2 + 2n + 1
= (n + 1)2
![Page 149: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/149.jpg)
Sum of the Odd Integers is a Square
n∑k=1
(2k − 1) = 1 + 3 + 5 + · · ·+ (2n − 1) = n2
Induction: If∑n
k=1(2k − 1) = n2, then
n+1∑k=1
(2k − 1) =
[n∑
k=1
(2k − 1)
]+ 2(n + 1)− 1
= n2 + 2(n + 1)− 1
= n2 + 2n + 1
= (n + 1)2
![Page 150: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/150.jpg)
Sum of the Odd Integers is a Square
n∑k=1
(2k − 1) = 1 + 3 + 5 + · · ·+ (2n − 1) = n2
Induction: If∑n
k=1(2k − 1) = n2, then
n+1∑k=1
(2k − 1) =
[n∑
k=1
(2k − 1)
]+ 2(n + 1)− 1
= n2 + 2(n + 1)− 1
= n2 + 2n + 1
= (n + 1)2
![Page 151: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/151.jpg)
Sum of the Odd Integers is a Square
n∑k=1
(2k − 1) = 1 + 3 + 5 + · · ·+ (2n − 1) = n2
Induction: If∑n
k=1(2k − 1) = n2, then
n+1∑k=1
(2k − 1) =
[n∑
k=1
(2k − 1)
]+ 2(n + 1)− 1
= n2 + 2(n + 1)− 1
= n2 + 2n + 1
= (n + 1)2
![Page 152: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/152.jpg)
Sum of the Odd Integers is a Square
n∑k=1
(2k − 1) = 1 + 3 + 5 + · · ·+ (2n − 1) = n2
Induction: If∑n
k=1(2k − 1) = n2, then
n+1∑k=1
(2k − 1) =
[n∑
k=1
(2k − 1)
]+ 2(n + 1)− 1
= n2 + 2(n + 1)− 1
= n2 + 2n + 1
= (n + 1)2
![Page 153: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/153.jpg)
Sum of Cubes is a Squared Triangular Number
The formulan−1∑k=1
k3 =
(12n(n − 1)
)2
can be established by mathematical induction
Basis: For n = 0 the sum on the left is empty, and so equalto 0. And the right hand side n is 0 too.
![Page 154: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/154.jpg)
Sum of Cubes is a Squared Triangular Number
The formulan−1∑k=1
k3 =
(12n(n − 1)
)2
can be established by mathematical induction
Basis: For n = 0 the sum on the left is empty, and so equalto 0. And the right hand side n is 0 too.
![Page 155: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/155.jpg)
Sum of Cubes is a Squared Triangular Number
The formulan−1∑k=1
k3 =
(12n(n − 1)
)2
can be established by mathematical induction
Basis: For n = 0 the sum on the left is empty, and so equalto 0. And the right hand side n is 0 too.
![Page 156: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/156.jpg)
Sum of Cubes is a Squared Triangular Number
n−1∑k=1
k3 =
(12n(n − 1)
)2
Induction: If∑n−1
k=1 k3 = n2(n − 1)2/4, then
n∑k=1
k3 =
[n−1∑k=1
k3
]+ n3
=
(12n(n − 1)
)2
+ n3
= n2(
(n − 1)2
4+ n)
= n2(
(n + 2n + 1)4
)=
(12n(n + 1)
)2
![Page 157: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/157.jpg)
Sum of Cubes is a Squared Triangular Number
n−1∑k=1
k3 =
(12n(n − 1)
)2
Induction: If∑n−1
k=1 k3 = n2(n − 1)2/4, then
n∑k=1
k3 =
[n−1∑k=1
k3
]+ n3
=
(12n(n − 1)
)2
+ n3
= n2(
(n − 1)2
4+ n)
= n2(
(n + 2n + 1)4
)=
(12n(n + 1)
)2
![Page 158: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/158.jpg)
Sum of Cubes is a Squared Triangular Number
n−1∑k=1
k3 =
(12n(n − 1)
)2
Induction: If∑n−1
k=1 k3 = n2(n − 1)2/4, then
n∑k=1
k3 =
[n−1∑k=1
k3
]+ n3
=
(12n(n − 1)
)2
+ n3
= n2(
(n − 1)2
4+ n)
= n2(
(n + 2n + 1)4
)=
(12n(n + 1)
)2
![Page 159: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/159.jpg)
Sum of Cubes is a Squared Triangular Number
n−1∑k=1
k3 =
(12n(n − 1)
)2
Induction: If∑n−1
k=1 k3 = n2(n − 1)2/4, then
n∑k=1
k3
=
[n−1∑k=1
k3
]+ n3
=
(12n(n − 1)
)2
+ n3
= n2(
(n − 1)2
4+ n)
= n2(
(n + 2n + 1)4
)=
(12n(n + 1)
)2
![Page 160: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/160.jpg)
Sum of Cubes is a Squared Triangular Number
n−1∑k=1
k3 =
(12n(n − 1)
)2
Induction: If∑n−1
k=1 k3 = n2(n − 1)2/4, then
n∑k=1
k3 =
[n−1∑k=1
k3
]+ n3
=
(12n(n − 1)
)2
+ n3
= n2(
(n − 1)2
4+ n)
= n2(
(n + 2n + 1)4
)=
(12n(n + 1)
)2
![Page 161: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/161.jpg)
Sum of Cubes is a Squared Triangular Number
n−1∑k=1
k3 =
(12n(n − 1)
)2
Induction: If∑n−1
k=1 k3 = n2(n − 1)2/4, then
n∑k=1
k3 =
[n−1∑k=1
k3
]+ n3
=
(12n(n − 1)
)2
+ n3
= n2(
(n − 1)2
4+ n)
= n2(
(n + 2n + 1)4
)=
(12n(n + 1)
)2
![Page 162: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/162.jpg)
Sum of Cubes is a Squared Triangular Number
n−1∑k=1
k3 =
(12n(n − 1)
)2
Induction: If∑n−1
k=1 k3 = n2(n − 1)2/4, then
n∑k=1
k3 =
[n−1∑k=1
k3
]+ n3
=
(12n(n − 1)
)2
+ n3
= n2(
(n − 1)2
4+ n)
= n2(
(n + 2n + 1)4
)=
(12n(n + 1)
)2
![Page 163: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/163.jpg)
Sum of Cubes is a Squared Triangular Number
n−1∑k=1
k3 =
(12n(n − 1)
)2
Induction: If∑n−1
k=1 k3 = n2(n − 1)2/4, then
n∑k=1
k3 =
[n−1∑k=1
k3
]+ n3
=
(12n(n − 1)
)2
+ n3
= n2(
(n − 1)2
4+ n)
= n2(
(n + 2n + 1)4
)
=
(12n(n + 1)
)2
![Page 164: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/164.jpg)
Sum of Cubes is a Squared Triangular Number
n−1∑k=1
k3 =
(12n(n − 1)
)2
Induction: If∑n−1
k=1 k3 = n2(n − 1)2/4, then
n∑k=1
k3 =
[n−1∑k=1
k3
]+ n3
=
(12n(n − 1)
)2
+ n3
= n2(
(n − 1)2
4+ n)
= n2(
(n + 2n + 1)4
)=
(12n(n + 1)
)2
![Page 165: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/165.jpg)
Sum of Fibonacci Numbers
Recall the Fibonacci sequence
~F = 〈0, 1, 1, 2, 3, 5, 8, 13, 21, . . .〉
Also recall the recurrence for the Fibonacci numbers
fn = fn−1 + fn−2, n ∈ {2, 3, 4, . . .}, f0 = 0, f1 = 1
Let’s write the sequence using the names of the numbersrather than their values
~F = 〈f0, f1, f2, f3, f4, f5, f6, f7, f8, . . .〉
![Page 166: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/166.jpg)
Sum of Fibonacci Numbers
Recall the Fibonacci sequence
~F = 〈0, 1, 1, 2, 3, 5, 8, 13, 21, . . .〉
Also recall the recurrence for the Fibonacci numbers
fn = fn−1 + fn−2, n ∈ {2, 3, 4, . . .}, f0 = 0, f1 = 1
Let’s write the sequence using the names of the numbersrather than their values
~F = 〈f0, f1, f2, f3, f4, f5, f6, f7, f8, . . .〉
![Page 167: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/167.jpg)
Sum of Fibonacci Numbers
Recall the Fibonacci sequence
~F = 〈0, 1, 1, 2, 3, 5, 8, 13, 21, . . .〉
Also recall the recurrence for the Fibonacci numbers
fn = fn−1 + fn−2, n ∈ {2, 3, 4, . . .}, f0 = 0, f1 = 1
Let’s write the sequence using the names of the numbersrather than their values
~F = 〈f0, f1, f2, f3, f4, f5, f6, f7, f8, . . .〉
![Page 168: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/168.jpg)
Sum of Fibonacci Numbers
Recall the Fibonacci sequence
~F = 〈0, 1, 1, 2, 3, 5, 8, 13, 21, . . .〉
Also recall the recurrence for the Fibonacci numbers
fn = fn−1 + fn−2, n ∈ {2, 3, 4, . . .}, f0 = 0, f1 = 1
Let’s write the sequence using the names of the numbersrather than their values
~F = 〈f0, f1, f2, f3, f4, f5, f6, f7, f8, . . .〉
![Page 169: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/169.jpg)
Sum of Fibonacci Numbers
~F = 〈0, 1, 1, 2, 3, 5, 8, 13, 21, . . .〉~F = 〈f0, f1, f2, f3, f4, f5, f6, f7, f8, . . .〉
Consider the sequence of partial sums
S0 = f0 = 0S1 = f0 + f1 = 1S2 = f0 + f1 + f2 = 2S3 = f0 + f1 + f2 + f3 = 4S4 = f0 + f1 + f2 + f3 + f4 = 7S4 = f0 + f1 + f2 + f3 + f4 + f5 = 12
Can you make a conjecture about these sums?
![Page 170: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/170.jpg)
Sum of Fibonacci Numbers
~F = 〈0, 1, 1, 2, 3, 5, 8, 13, 21, . . .〉~F = 〈f0, f1, f2, f3, f4, f5, f6, f7, f8, . . .〉
Consider the sequence of partial sums
S0 = f0 = 0S1 = f0 + f1 = 1S2 = f0 + f1 + f2 = 2S3 = f0 + f1 + f2 + f3 = 4S4 = f0 + f1 + f2 + f3 + f4 = 7S4 = f0 + f1 + f2 + f3 + f4 + f5 = 12
Can you make a conjecture about these sums?
![Page 171: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/171.jpg)
Sum of Fibonacci Numbers
~F = 〈0, 1, 1, 2, 3, 5, 8, 13, 21, . . .〉~F = 〈f0, f1, f2, f3, f4, f5, f6, f7, f8, . . .〉
Consider the sequence of partial sums
S0 = f0 = 0S1 = f0 + f1 = 1S2 = f0 + f1 + f2 = 2S3 = f0 + f1 + f2 + f3 = 4S4 = f0 + f1 + f2 + f3 + f4 = 7S4 = f0 + f1 + f2 + f3 + f4 + f5 = 12
Can you make a conjecture about these sums?
![Page 172: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/172.jpg)
Sum of Fibonacci Numbers
~F = 〈0, 1, 1, 2, 3, 5, 8, 13, 21, . . .〉~F = 〈f0, f1, f2, f3, f4, f5, f6, f7, f8, . . .〉
Consider the sequence of partial sums
S0
= f0 = 0S1 = f0 + f1 = 1S2 = f0 + f1 + f2 = 2S3 = f0 + f1 + f2 + f3 = 4S4 = f0 + f1 + f2 + f3 + f4 = 7S4 = f0 + f1 + f2 + f3 + f4 + f5 = 12
Can you make a conjecture about these sums?
![Page 173: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/173.jpg)
Sum of Fibonacci Numbers
~F = 〈0, 1, 1, 2, 3, 5, 8, 13, 21, . . .〉~F = 〈f0, f1, f2, f3, f4, f5, f6, f7, f8, . . .〉
Consider the sequence of partial sums
S0 = f0 = 0
S1 = f0 + f1 = 1S2 = f0 + f1 + f2 = 2S3 = f0 + f1 + f2 + f3 = 4S4 = f0 + f1 + f2 + f3 + f4 = 7S4 = f0 + f1 + f2 + f3 + f4 + f5 = 12
Can you make a conjecture about these sums?
![Page 174: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/174.jpg)
Sum of Fibonacci Numbers
~F = 〈0, 1, 1, 2, 3, 5, 8, 13, 21, . . .〉~F = 〈f0, f1, f2, f3, f4, f5, f6, f7, f8, . . .〉
Consider the sequence of partial sums
S0 = f0 = 0S1
= f0 + f1 = 1S2 = f0 + f1 + f2 = 2S3 = f0 + f1 + f2 + f3 = 4S4 = f0 + f1 + f2 + f3 + f4 = 7S4 = f0 + f1 + f2 + f3 + f4 + f5 = 12
Can you make a conjecture about these sums?
![Page 175: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/175.jpg)
Sum of Fibonacci Numbers
~F = 〈0, 1, 1, 2, 3, 5, 8, 13, 21, . . .〉~F = 〈f0, f1, f2, f3, f4, f5, f6, f7, f8, . . .〉
Consider the sequence of partial sums
S0 = f0 = 0S1 = f0 + f1 = 1
S2 = f0 + f1 + f2 = 2S3 = f0 + f1 + f2 + f3 = 4S4 = f0 + f1 + f2 + f3 + f4 = 7S4 = f0 + f1 + f2 + f3 + f4 + f5 = 12
Can you make a conjecture about these sums?
![Page 176: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/176.jpg)
Sum of Fibonacci Numbers
~F = 〈0, 1, 1, 2, 3, 5, 8, 13, 21, . . .〉~F = 〈f0, f1, f2, f3, f4, f5, f6, f7, f8, . . .〉
Consider the sequence of partial sums
S0 = f0 = 0S1 = f0 + f1 = 1S2
= f0 + f1 + f2 = 2S3 = f0 + f1 + f2 + f3 = 4S4 = f0 + f1 + f2 + f3 + f4 = 7S4 = f0 + f1 + f2 + f3 + f4 + f5 = 12
Can you make a conjecture about these sums?
![Page 177: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/177.jpg)
Sum of Fibonacci Numbers
~F = 〈0, 1, 1, 2, 3, 5, 8, 13, 21, . . .〉~F = 〈f0, f1, f2, f3, f4, f5, f6, f7, f8, . . .〉
Consider the sequence of partial sums
S0 = f0 = 0S1 = f0 + f1 = 1S2 = f0 + f1 + f2 = 2
S3 = f0 + f1 + f2 + f3 = 4S4 = f0 + f1 + f2 + f3 + f4 = 7S4 = f0 + f1 + f2 + f3 + f4 + f5 = 12
Can you make a conjecture about these sums?
![Page 178: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/178.jpg)
Sum of Fibonacci Numbers
~F = 〈0, 1, 1, 2, 3, 5, 8, 13, 21, . . .〉~F = 〈f0, f1, f2, f3, f4, f5, f6, f7, f8, . . .〉
Consider the sequence of partial sums
S0 = f0 = 0S1 = f0 + f1 = 1S2 = f0 + f1 + f2 = 2S3
= f0 + f1 + f2 + f3 = 4S4 = f0 + f1 + f2 + f3 + f4 = 7S4 = f0 + f1 + f2 + f3 + f4 + f5 = 12
Can you make a conjecture about these sums?
![Page 179: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/179.jpg)
Sum of Fibonacci Numbers
~F = 〈0, 1, 1, 2, 3, 5, 8, 13, 21, . . .〉~F = 〈f0, f1, f2, f3, f4, f5, f6, f7, f8, . . .〉
Consider the sequence of partial sums
S0 = f0 = 0S1 = f0 + f1 = 1S2 = f0 + f1 + f2 = 2S3 = f0 + f1 + f2 + f3 = 4
S4 = f0 + f1 + f2 + f3 + f4 = 7S4 = f0 + f1 + f2 + f3 + f4 + f5 = 12
Can you make a conjecture about these sums?
![Page 180: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/180.jpg)
Sum of Fibonacci Numbers
~F = 〈0, 1, 1, 2, 3, 5, 8, 13, 21, . . .〉~F = 〈f0, f1, f2, f3, f4, f5, f6, f7, f8, . . .〉
Consider the sequence of partial sums
S0 = f0 = 0S1 = f0 + f1 = 1S2 = f0 + f1 + f2 = 2S3 = f0 + f1 + f2 + f3 = 4S4
= f0 + f1 + f2 + f3 + f4 = 7S4 = f0 + f1 + f2 + f3 + f4 + f5 = 12
Can you make a conjecture about these sums?
![Page 181: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/181.jpg)
Sum of Fibonacci Numbers
~F = 〈0, 1, 1, 2, 3, 5, 8, 13, 21, . . .〉~F = 〈f0, f1, f2, f3, f4, f5, f6, f7, f8, . . .〉
Consider the sequence of partial sums
S0 = f0 = 0S1 = f0 + f1 = 1S2 = f0 + f1 + f2 = 2S3 = f0 + f1 + f2 + f3 = 4S4 = f0 + f1 + f2 + f3 + f4 = 7
S4 = f0 + f1 + f2 + f3 + f4 + f5 = 12
Can you make a conjecture about these sums?
![Page 182: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/182.jpg)
Sum of Fibonacci Numbers
~F = 〈0, 1, 1, 2, 3, 5, 8, 13, 21, . . .〉~F = 〈f0, f1, f2, f3, f4, f5, f6, f7, f8, . . .〉
Consider the sequence of partial sums
S0 = f0 = 0S1 = f0 + f1 = 1S2 = f0 + f1 + f2 = 2S3 = f0 + f1 + f2 + f3 = 4S4 = f0 + f1 + f2 + f3 + f4 = 7S4
= f0 + f1 + f2 + f3 + f4 + f5 = 12
Can you make a conjecture about these sums?
![Page 183: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/183.jpg)
Sum of Fibonacci Numbers
~F = 〈0, 1, 1, 2, 3, 5, 8, 13, 21, . . .〉~F = 〈f0, f1, f2, f3, f4, f5, f6, f7, f8, . . .〉
Consider the sequence of partial sums
S0 = f0 = 0S1 = f0 + f1 = 1S2 = f0 + f1 + f2 = 2S3 = f0 + f1 + f2 + f3 = 4S4 = f0 + f1 + f2 + f3 + f4 = 7S4 = f0 + f1 + f2 + f3 + f4 + f5 = 12
Can you make a conjecture about these sums?
![Page 184: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/184.jpg)
Sum of Fibonacci Numbers
n−1∑k=0
fk = fn+1 − 1
Basis: For n = 0
The sum on the left-hand side is empty and soequal to 0The formula on the right-hand side is also equalto 0
f0+1 − 1 = f1 − 1 = 1− 1 = 0
This establishes the basis for induction
![Page 185: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/185.jpg)
Sum of Fibonacci Numbers
n−1∑k=0
fk = fn+1 − 1
Basis: For n = 0
The sum on the left-hand side is empty and soequal to 0The formula on the right-hand side is also equalto 0
f0+1 − 1 = f1 − 1 = 1− 1 = 0
This establishes the basis for induction
![Page 186: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/186.jpg)
Sum of Fibonacci Numbers
n−1∑k=0
fk = fn+1 − 1
Basis: For n = 0The sum on the left-hand side is empty and soequal to 0
The formula on the right-hand side is also equalto 0
f0+1 − 1 = f1 − 1 = 1− 1 = 0
This establishes the basis for induction
![Page 187: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/187.jpg)
Sum of Fibonacci Numbers
n−1∑k=0
fk = fn+1 − 1
Basis: For n = 0The sum on the left-hand side is empty and soequal to 0The formula on the right-hand side is also equalto 0
f0+1 − 1 = f1 − 1 = 1− 1 = 0
This establishes the basis for induction
![Page 188: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/188.jpg)
Sum of Fibonacci Numbers
n−1∑k=0
fk = fn+1 − 1
Basis: For n = 0The sum on the left-hand side is empty and soequal to 0The formula on the right-hand side is also equalto 0
f0+1 − 1 = f1 − 1 = 1− 1 = 0
This establishes the basis for induction
![Page 189: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/189.jpg)
Sum of Fibonacci Numbers
n−1∑k=0
fk = fn+1 − 1
Induction: Pretend thatn−1∑k=0
fk = fn+1 − 1
is true for some n ≥ 0
Then
n∑k=0
fk =
[n−1∑k=1
fk
]+ fn
= [fn+1 − 1] + fn= [fn+1 + fn]− 1= fn+2 − 1
![Page 190: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/190.jpg)
Sum of Fibonacci Numbers
n−1∑k=0
fk = fn+1 − 1
Induction: Pretend thatn−1∑k=0
fk = fn+1 − 1
is true for some n ≥ 0
Then
n∑k=0
fk =
[n−1∑k=1
fk
]+ fn
= [fn+1 − 1] + fn= [fn+1 + fn]− 1= fn+2 − 1
![Page 191: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/191.jpg)
Sum of Fibonacci Numbers
n−1∑k=0
fk = fn+1 − 1
Induction: Pretend thatn−1∑k=0
fk = fn+1 − 1
is true for some n ≥ 0Then
n∑k=0
fk =
[n−1∑k=1
fk
]+ fn
= [fn+1 − 1] + fn= [fn+1 + fn]− 1= fn+2 − 1
![Page 192: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/192.jpg)
Sum of Fibonacci Numbers
n−1∑k=0
fk = fn+1 − 1
Induction: Pretend thatn−1∑k=0
fk = fn+1 − 1
is true for some n ≥ 0Then
n∑k=0
fk =
[n−1∑k=1
fk
]+ fn
= [fn+1 − 1] + fn= [fn+1 + fn]− 1= fn+2 − 1
![Page 193: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/193.jpg)
Sum of Fibonacci Numbers
n−1∑k=0
fk = fn+1 − 1
Induction: Pretend thatn−1∑k=0
fk = fn+1 − 1
is true for some n ≥ 0Then
n∑k=0
fk
=
[n−1∑k=1
fk
]+ fn
= [fn+1 − 1] + fn= [fn+1 + fn]− 1= fn+2 − 1
![Page 194: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/194.jpg)
Sum of Fibonacci Numbers
n−1∑k=0
fk = fn+1 − 1
Induction: Pretend thatn−1∑k=0
fk = fn+1 − 1
is true for some n ≥ 0Then
n∑k=0
fk =
[n−1∑k=1
fk
]+ fn
= [fn+1 − 1] + fn= [fn+1 + fn]− 1= fn+2 − 1
![Page 195: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/195.jpg)
Sum of Fibonacci Numbers
n−1∑k=0
fk = fn+1 − 1
Induction: Pretend thatn−1∑k=0
fk = fn+1 − 1
is true for some n ≥ 0Then
n∑k=0
fk =
[n−1∑k=1
fk
]+ fn
= [fn+1 − 1] + fn
= [fn+1 + fn]− 1= fn+2 − 1
![Page 196: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/196.jpg)
Sum of Fibonacci Numbers
n−1∑k=0
fk = fn+1 − 1
Induction: Pretend thatn−1∑k=0
fk = fn+1 − 1
is true for some n ≥ 0Then
n∑k=0
fk =
[n−1∑k=1
fk
]+ fn
= [fn+1 − 1] + fn= [fn+1 + fn]− 1
= fn+2 − 1
![Page 197: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/197.jpg)
Sum of Fibonacci Numbers
n−1∑k=0
fk = fn+1 − 1
Induction: Pretend thatn−1∑k=0
fk = fn+1 − 1
is true for some n ≥ 0Then
n∑k=0
fk =
[n−1∑k=1
fk
]+ fn
= [fn+1 − 1] + fn= [fn+1 + fn]− 1= fn+2 − 1
![Page 198: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/198.jpg)
Pascal’s Triangle
11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 11 7 21 35 35 21 7 11 8 28 56 70 56 28 8 1...
......
......
......
......
. . .
![Page 199: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/199.jpg)
Pascal’s Triangle
11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 11 7 21 35 35 21 7 11 8 28 56 70 56 28 8 1...
......
......
......
......
. . .
![Page 200: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/200.jpg)
Pascal’s Triangle: Sum of Rows
There are many identities that can be found in Pascal’s trianglePerhaps the most famous one is found by summing values in arow
1 = 11 +1 = 21 +2 +1 = 41 +3 +3 +1 = 81 +4 +6 +4 +1 = 161 +5 +10 +10 +5 +1 = 321 +6 +15 +20 +15 +6 +1 = 64
![Page 201: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/201.jpg)
Pascal’s Triangle: Sum of Rows
There are many identities that can be found in Pascal’s triangle
Perhaps the most famous one is found by summing values in arow
1 = 11 +1 = 21 +2 +1 = 41 +3 +3 +1 = 81 +4 +6 +4 +1 = 161 +5 +10 +10 +5 +1 = 321 +6 +15 +20 +15 +6 +1 = 64
![Page 202: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/202.jpg)
Pascal’s Triangle: Sum of Rows
There are many identities that can be found in Pascal’s trianglePerhaps the most famous one is found by summing values in arow
1 = 11 +1 = 21 +2 +1 = 41 +3 +3 +1 = 81 +4 +6 +4 +1 = 161 +5 +10 +10 +5 +1 = 321 +6 +15 +20 +15 +6 +1 = 64
![Page 203: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/203.jpg)
Pascal’s Triangle: Sum of Rows
There are many identities that can be found in Pascal’s trianglePerhaps the most famous one is found by summing values in arow
1 = 11 +1 = 21 +2 +1 = 41 +3 +3 +1 = 81 +4 +6 +4 +1 = 161 +5 +10 +10 +5 +1 = 321 +6 +15 +20 +15 +6 +1 = 64
![Page 204: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/204.jpg)
Pascal’s Triangle: Sum of Rows
There are many identities that can be found in Pascal’s trianglePerhaps the most famous one is found by summing values in arow
1
= 11 +1 = 21 +2 +1 = 41 +3 +3 +1 = 81 +4 +6 +4 +1 = 161 +5 +10 +10 +5 +1 = 321 +6 +15 +20 +15 +6 +1 = 64
![Page 205: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/205.jpg)
Pascal’s Triangle: Sum of Rows
There are many identities that can be found in Pascal’s trianglePerhaps the most famous one is found by summing values in arow
1 = 1
1 +1 = 21 +2 +1 = 41 +3 +3 +1 = 81 +4 +6 +4 +1 = 161 +5 +10 +10 +5 +1 = 321 +6 +15 +20 +15 +6 +1 = 64
![Page 206: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/206.jpg)
Pascal’s Triangle: Sum of Rows
There are many identities that can be found in Pascal’s trianglePerhaps the most famous one is found by summing values in arow
1 = 11 +1
= 21 +2 +1 = 41 +3 +3 +1 = 81 +4 +6 +4 +1 = 161 +5 +10 +10 +5 +1 = 321 +6 +15 +20 +15 +6 +1 = 64
![Page 207: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/207.jpg)
Pascal’s Triangle: Sum of Rows
There are many identities that can be found in Pascal’s trianglePerhaps the most famous one is found by summing values in arow
1 = 11 +1 = 2
1 +2 +1 = 41 +3 +3 +1 = 81 +4 +6 +4 +1 = 161 +5 +10 +10 +5 +1 = 321 +6 +15 +20 +15 +6 +1 = 64
![Page 208: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/208.jpg)
Pascal’s Triangle: Sum of Rows
There are many identities that can be found in Pascal’s trianglePerhaps the most famous one is found by summing values in arow
1 = 11 +1 = 21 +2 +1
= 41 +3 +3 +1 = 81 +4 +6 +4 +1 = 161 +5 +10 +10 +5 +1 = 321 +6 +15 +20 +15 +6 +1 = 64
![Page 209: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/209.jpg)
Pascal’s Triangle: Sum of Rows
There are many identities that can be found in Pascal’s trianglePerhaps the most famous one is found by summing values in arow
1 = 11 +1 = 21 +2 +1 = 4
1 +3 +3 +1 = 81 +4 +6 +4 +1 = 161 +5 +10 +10 +5 +1 = 321 +6 +15 +20 +15 +6 +1 = 64
![Page 210: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/210.jpg)
Pascal’s Triangle: Sum of Rows
There are many identities that can be found in Pascal’s trianglePerhaps the most famous one is found by summing values in arow
1 = 11 +1 = 21 +2 +1 = 41 +3 +3 +1
= 81 +4 +6 +4 +1 = 161 +5 +10 +10 +5 +1 = 321 +6 +15 +20 +15 +6 +1 = 64
![Page 211: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/211.jpg)
Pascal’s Triangle: Sum of Rows
There are many identities that can be found in Pascal’s trianglePerhaps the most famous one is found by summing values in arow
1 = 11 +1 = 21 +2 +1 = 41 +3 +3 +1 = 8
1 +4 +6 +4 +1 = 161 +5 +10 +10 +5 +1 = 321 +6 +15 +20 +15 +6 +1 = 64
![Page 212: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/212.jpg)
Pascal’s Triangle: Sum of Rows
There are many identities that can be found in Pascal’s trianglePerhaps the most famous one is found by summing values in arow
1 = 11 +1 = 21 +2 +1 = 41 +3 +3 +1 = 81 +4 +6 +4 +1
= 161 +5 +10 +10 +5 +1 = 321 +6 +15 +20 +15 +6 +1 = 64
![Page 213: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/213.jpg)
Pascal’s Triangle: Sum of Rows
There are many identities that can be found in Pascal’s trianglePerhaps the most famous one is found by summing values in arow
1 = 11 +1 = 21 +2 +1 = 41 +3 +3 +1 = 81 +4 +6 +4 +1 = 16
1 +5 +10 +10 +5 +1 = 321 +6 +15 +20 +15 +6 +1 = 64
![Page 214: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/214.jpg)
Pascal’s Triangle: Sum of Rows
There are many identities that can be found in Pascal’s trianglePerhaps the most famous one is found by summing values in arow
1 = 11 +1 = 21 +2 +1 = 41 +3 +3 +1 = 81 +4 +6 +4 +1 = 161 +5 +10 +10 +5 +1
= 321 +6 +15 +20 +15 +6 +1 = 64
![Page 215: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/215.jpg)
Pascal’s Triangle: Sum of Rows
There are many identities that can be found in Pascal’s trianglePerhaps the most famous one is found by summing values in arow
1 = 11 +1 = 21 +2 +1 = 41 +3 +3 +1 = 81 +4 +6 +4 +1 = 161 +5 +10 +10 +5 +1 = 32
1 +6 +15 +20 +15 +6 +1 = 64
![Page 216: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/216.jpg)
Pascal’s Triangle: Sum of Rows
There are many identities that can be found in Pascal’s trianglePerhaps the most famous one is found by summing values in arow
1 = 11 +1 = 21 +2 +1 = 41 +3 +3 +1 = 81 +4 +6 +4 +1 = 161 +5 +10 +10 +5 +1 = 321 +6 +15 +20 +15 +6 +1
= 64
![Page 217: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/217.jpg)
Pascal’s Triangle: Sum of Rows
There are many identities that can be found in Pascal’s trianglePerhaps the most famous one is found by summing values in arow
1 = 11 +1 = 21 +2 +1 = 41 +3 +3 +1 = 81 +4 +6 +4 +1 = 161 +5 +10 +10 +5 +1 = 321 +6 +15 +20 +15 +6 +1 = 64
![Page 218: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/218.jpg)
Pascal’s Triangle: Sum of Rows
We can make the conjecture
The sum of values in row n of Pascal’s triangle is 2n forn = 0, 1, 2, 3, . . .
When you notice how a row is computed from the previousrow, it becomes clear why the row sums doubleEvery term, except the first and last 1, is added twice togenerate the next row
![Page 219: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/219.jpg)
Pascal’s Triangle: Sum of Rows
We can make the conjecture
The sum of values in row n of Pascal’s triangle is 2n forn = 0, 1, 2, 3, . . .
When you notice how a row is computed from the previousrow, it becomes clear why the row sums doubleEvery term, except the first and last 1, is added twice togenerate the next row
![Page 220: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/220.jpg)
Pascal’s Triangle: Sum of Rows
We can make the conjectureThe sum of values in row n of Pascal’s triangle is 2n forn = 0, 1, 2, 3, . . .
When you notice how a row is computed from the previousrow, it becomes clear why the row sums doubleEvery term, except the first and last 1, is added twice togenerate the next row
![Page 221: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/221.jpg)
Pascal’s Triangle: Sum of Rows
We can make the conjectureThe sum of values in row n of Pascal’s triangle is 2n forn = 0, 1, 2, 3, . . .
When you notice how a row is computed from the previousrow, it becomes clear why the row sums double
Every term, except the first and last 1, is added twice togenerate the next row
![Page 222: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/222.jpg)
Pascal’s Triangle: Sum of Rows
We can make the conjectureThe sum of values in row n of Pascal’s triangle is 2n forn = 0, 1, 2, 3, . . .
When you notice how a row is computed from the previousrow, it becomes clear why the row sums doubleEvery term, except the first and last 1, is added twice togenerate the next row
![Page 223: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/223.jpg)
Pascal’s Triangle: Sum of Rows
Consider row 6 of Pascal’s Triangle
1 6 15 20 15 6 1
It can be used to generate row 7
1 7 21 35 35 21 7 1
Sum adjacent numbers in row 6 yields
1 1 + 6 6 + 15 15 + 20 20 + 15 15 + 6 6 + 1 11 7 21 35 35 21 7 1
Adding in the initial and terminal 1 in row 7 shows each termin row 6 is summed twice when summing row 7
![Page 224: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/224.jpg)
Pascal’s Triangle: Sum of Rows
Consider row 6 of Pascal’s Triangle
1 6 15 20 15 6 1
It can be used to generate row 7
1 7 21 35 35 21 7 1
Sum adjacent numbers in row 6 yields
1 1 + 6 6 + 15 15 + 20 20 + 15 15 + 6 6 + 1 11 7 21 35 35 21 7 1
Adding in the initial and terminal 1 in row 7 shows each termin row 6 is summed twice when summing row 7
![Page 225: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/225.jpg)
Pascal’s Triangle: Sum of Rows
Consider row 6 of Pascal’s Triangle
1 6 15 20 15 6 1
It can be used to generate row 7
1 7 21 35 35 21 7 1
Sum adjacent numbers in row 6 yields
1 1 + 6 6 + 15 15 + 20 20 + 15 15 + 6 6 + 1 11 7 21 35 35 21 7 1
Adding in the initial and terminal 1 in row 7 shows each termin row 6 is summed twice when summing row 7
![Page 226: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/226.jpg)
Pascal’s Triangle: Sum of Rows
Consider row 6 of Pascal’s Triangle
1 6 15 20 15 6 1
It can be used to generate row 7
1 7 21 35 35 21 7 1
Sum adjacent numbers in row 6 yields
1 1 + 6 6 + 15 15 + 20 20 + 15 15 + 6 6 + 1 11 7 21 35 35 21 7 1
Adding in the initial and terminal 1 in row 7 shows each termin row 6 is summed twice when summing row 7
![Page 227: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/227.jpg)
Pascal’s Triangle: Sum of Rows
Consider row 6 of Pascal’s Triangle
1 6 15 20 15 6 1
It can be used to generate row 7
1 7 21 35 35 21 7 1
Sum adjacent numbers in row 6 yields
1
1 + 6
6 + 15 15 + 20 20 + 15 15 + 6 6 + 1 11 7 21 35 35 21 7 1
Adding in the initial and terminal 1 in row 7 shows each termin row 6 is summed twice when summing row 7
![Page 228: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/228.jpg)
Pascal’s Triangle: Sum of Rows
Consider row 6 of Pascal’s Triangle
1 6 15 20 15 6 1
It can be used to generate row 7
1 7 21 35 35 21 7 1
Sum adjacent numbers in row 6 yields
1
1 + 6
6 + 15 15 + 20 20 + 15 15 + 6 6 + 1 11
7
21 35 35 21 7 1
Adding in the initial and terminal 1 in row 7 shows each termin row 6 is summed twice when summing row 7
![Page 229: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/229.jpg)
Pascal’s Triangle: Sum of Rows
Consider row 6 of Pascal’s Triangle
1 6 15 20 15 6 1
It can be used to generate row 7
1 7 21 35 35 21 7 1
Sum adjacent numbers in row 6 yields
1
1 + 6 6 + 15
15 + 20 20 + 15 15 + 6 6 + 1 11
7
21 35 35 21 7 1
Adding in the initial and terminal 1 in row 7 shows each termin row 6 is summed twice when summing row 7
![Page 230: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/230.jpg)
Pascal’s Triangle: Sum of Rows
Consider row 6 of Pascal’s Triangle
1 6 15 20 15 6 1
It can be used to generate row 7
1 7 21 35 35 21 7 1
Sum adjacent numbers in row 6 yields
1
1 + 6 6 + 15
15 + 20 20 + 15 15 + 6 6 + 1 11
7 21
35 35 21 7 1
Adding in the initial and terminal 1 in row 7 shows each termin row 6 is summed twice when summing row 7
![Page 231: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/231.jpg)
Pascal’s Triangle: Sum of Rows
Consider row 6 of Pascal’s Triangle
1 6 15 20 15 6 1
It can be used to generate row 7
1 7 21 35 35 21 7 1
Sum adjacent numbers in row 6 yields
1
1 + 6 6 + 15 15 + 20
20 + 15 15 + 6 6 + 1 11
7 21
35 35 21 7 1
Adding in the initial and terminal 1 in row 7 shows each termin row 6 is summed twice when summing row 7
![Page 232: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/232.jpg)
Pascal’s Triangle: Sum of Rows
Consider row 6 of Pascal’s Triangle
1 6 15 20 15 6 1
It can be used to generate row 7
1 7 21 35 35 21 7 1
Sum adjacent numbers in row 6 yields
1
1 + 6 6 + 15 15 + 20
20 + 15 15 + 6 6 + 1 11
7 21 35
35 21 7 1
Adding in the initial and terminal 1 in row 7 shows each termin row 6 is summed twice when summing row 7
![Page 233: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/233.jpg)
Pascal’s Triangle: Sum of Rows
Consider row 6 of Pascal’s Triangle
1 6 15 20 15 6 1
It can be used to generate row 7
1 7 21 35 35 21 7 1
Sum adjacent numbers in row 6 yields
1
1 + 6 6 + 15 15 + 20 20 + 15
15 + 6 6 + 1 11
7 21 35
35 21 7 1
Adding in the initial and terminal 1 in row 7 shows each termin row 6 is summed twice when summing row 7
![Page 234: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/234.jpg)
Pascal’s Triangle: Sum of Rows
Consider row 6 of Pascal’s Triangle
1 6 15 20 15 6 1
It can be used to generate row 7
1 7 21 35 35 21 7 1
Sum adjacent numbers in row 6 yields
1
1 + 6 6 + 15 15 + 20 20 + 15
15 + 6 6 + 1 11
7 21 35 35
21 7 1
Adding in the initial and terminal 1 in row 7 shows each termin row 6 is summed twice when summing row 7
![Page 235: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/235.jpg)
Pascal’s Triangle: Sum of Rows
Consider row 6 of Pascal’s Triangle
1 6 15 20 15 6 1
It can be used to generate row 7
1 7 21 35 35 21 7 1
Sum adjacent numbers in row 6 yields
1
1 + 6 6 + 15 15 + 20 20 + 15 15 + 6
6 + 1 11
7 21 35 35
21 7 1
Adding in the initial and terminal 1 in row 7 shows each termin row 6 is summed twice when summing row 7
![Page 236: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/236.jpg)
Pascal’s Triangle: Sum of Rows
Consider row 6 of Pascal’s Triangle
1 6 15 20 15 6 1
It can be used to generate row 7
1 7 21 35 35 21 7 1
Sum adjacent numbers in row 6 yields
1
1 + 6 6 + 15 15 + 20 20 + 15 15 + 6
6 + 1 11
7 21 35 35 21
7 1
Adding in the initial and terminal 1 in row 7 shows each termin row 6 is summed twice when summing row 7
![Page 237: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/237.jpg)
Pascal’s Triangle: Sum of Rows
Consider row 6 of Pascal’s Triangle
1 6 15 20 15 6 1
It can be used to generate row 7
1 7 21 35 35 21 7 1
Sum adjacent numbers in row 6 yields
1
1 + 6 6 + 15 15 + 20 20 + 15 15 + 6 6 + 1
11
7 21 35 35 21
7 1
Adding in the initial and terminal 1 in row 7 shows each termin row 6 is summed twice when summing row 7
![Page 238: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/238.jpg)
Pascal’s Triangle: Sum of Rows
Consider row 6 of Pascal’s Triangle
1 6 15 20 15 6 1
It can be used to generate row 7
1 7 21 35 35 21 7 1
Sum adjacent numbers in row 6 yields
1
1 + 6 6 + 15 15 + 20 20 + 15 15 + 6 6 + 1
11
7 21 35 35 21 7
1
Adding in the initial and terminal 1 in row 7 shows each termin row 6 is summed twice when summing row 7
![Page 239: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/239.jpg)
Pascal’s Triangle: Sum of Rows
Consider row 6 of Pascal’s Triangle
1 6 15 20 15 6 1
It can be used to generate row 7
1 7 21 35 35 21 7 1
Sum adjacent numbers in row 6 yields
1
1 + 6 6 + 15 15 + 20 20 + 15 15 + 6 6 + 1
11
7 21 35 35 21 7
1
Adding in the initial and terminal 1 in row 7 shows each termin row 6 is summed twice when summing row 7
![Page 240: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/240.jpg)
Pascal’s Triangle: Sum of Rows
Consider row 6 of Pascal’s Triangle
1 6 15 20 15 6 1
It can be used to generate row 7
1 7 21 35 35 21 7 1
Sum adjacent numbers in row 6 yields
1 1 + 6 6 + 15 15 + 20 20 + 15 15 + 6 6 + 1 11 7 21 35 35 21 7 1
Adding in the initial and terminal 1 in row 7 shows each termin row 6 is summed twice when summing row 7
![Page 241: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/241.jpg)
Pascal’s Triangle: Sum of Rows
To make the preceding argument precise we must name thingsWe’ll name the rows and columns by the natural numbers
0 1 2 3 4 5 6 7 8 · · ·0 11 1 12 1 2 13 1 3 3 14 1 4 6 4 15 1 5 10 10 5 16 1 6 15 20 15 6 17 1 7 21 35 35 21 7 18 1 8 28 56 70 56 28 8 1...
......
......
......
......
. . .
![Page 242: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/242.jpg)
Pascal’s Triangle: Sum of Rows
To make the preceding argument precise we must name things
We’ll name the rows and columns by the natural numbers
0 1 2 3 4 5 6 7 8 · · ·0 11 1 12 1 2 13 1 3 3 14 1 4 6 4 15 1 5 10 10 5 16 1 6 15 20 15 6 17 1 7 21 35 35 21 7 18 1 8 28 56 70 56 28 8 1...
......
......
......
......
. . .
![Page 243: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/243.jpg)
Pascal’s Triangle: Sum of Rows
To make the preceding argument precise we must name thingsWe’ll name the rows and columns by the natural numbers
0 1 2 3 4 5 6 7 8 · · ·0 11 1 12 1 2 13 1 3 3 14 1 4 6 4 15 1 5 10 10 5 16 1 6 15 20 15 6 17 1 7 21 35 35 21 7 18 1 8 28 56 70 56 28 8 1...
......
......
......
......
. . .
![Page 244: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/244.jpg)
Pascal’s Triangle: Sum of Rows
To make the preceding argument precise we must name thingsWe’ll name the rows and columns by the natural numbers
0 1 2 3 4 5 6 7 8 · · ·0 11 1 12 1 2 13 1 3 3 14 1 4 6 4 15 1 5 10 10 5 16 1 6 15 20 15 6 17 1 7 21 35 35 21 7 18 1 8 28 56 70 56 28 8 1...
......
......
......
......
. . .
![Page 245: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/245.jpg)
Pascal’s Triangle: Sum of Rows
To make the preceding argument precise we must name thingsWe’ll name the rows and columns by the natural numbers
0 1 2 3 4 5 6 7 8 · · ·0 11 1 12 1 2 13 1 3 3 14 1 4 6 4 15 1 5 10 10 5 16 1 6 15 20 15 6 17 1 7 21 35 35 21 7 18 1 8 28 56 70 56 28 8 1...
......
......
......
......
. . .
![Page 246: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/246.jpg)
Pascal’s Triangle: Binomial Coefficients
We’ll name terms in row n, column m by the symbol(nm
)The symbol
(nm
)is called a binomial coefficient
The symbol(nm
)is read “n choose m”
The binomial coefficient “n choose m” can be expressed interms of factorials (
nm
)=
n!m!(n −m)!
![Page 247: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/247.jpg)
Pascal’s Triangle: Binomial Coefficients
We’ll name terms in row n, column m by the symbol(nm
)
The symbol(nm
)is called a binomial coefficient
The symbol(nm
)is read “n choose m”
The binomial coefficient “n choose m” can be expressed interms of factorials (
nm
)=
n!m!(n −m)!
![Page 248: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/248.jpg)
Pascal’s Triangle: Binomial Coefficients
We’ll name terms in row n, column m by the symbol(nm
)The symbol
(nm
)is called a binomial coefficient
The symbol(nm
)is read “n choose m”
The binomial coefficient “n choose m” can be expressed interms of factorials (
nm
)=
n!m!(n −m)!
![Page 249: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/249.jpg)
Pascal’s Triangle: Binomial Coefficients
We’ll name terms in row n, column m by the symbol(nm
)The symbol
(nm
)is called a binomial coefficient
The symbol(nm
)is read “n choose m”
The binomial coefficient “n choose m” can be expressed interms of factorials (
nm
)=
n!m!(n −m)!
![Page 250: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/250.jpg)
Pascal’s Triangle: Binomial Coefficients
We’ll name terms in row n, column m by the symbol(nm
)The symbol
(nm
)is called a binomial coefficient
The symbol(nm
)is read “n choose m”
The binomial coefficient “n choose m” can be expressed interms of factorials (
nm
)=
n!m!(n −m)!
![Page 251: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/251.jpg)
Pascal’s Triangle
0 1 2 3 4 5 6 7 8 · · ·0(00
)1(10
) (11
)2(20
) (21
) (22
)3(30
) (31
) (32
) (33
)4(40
) (41
) (42
) (43
) (44
)5(50
) (51
) (52
) (53
) (54
) (55
)6(60
) (61
) (62
) (63
) (64
) (65
) (66
)7(70
) (71
) (72
) (73
) (74
) (75
) (76
) (77
)8(80
) (81
) (82
) (83
) (84
) (85
) (86
) (87
) (88
)...
......
......
......
......
. . .
![Page 252: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/252.jpg)
Pascal’s Triangle
0 1 2 3 4 5 6 7 8 · · ·0(00
)1(10
) (11
)2(20
) (21
) (22
)3(30
) (31
) (32
) (33
)4(40
) (41
) (42
) (43
) (44
)5(50
) (51
) (52
) (53
) (54
) (55
)6(60
) (61
) (62
) (63
) (64
) (65
) (66
)7(70
) (71
) (72
) (73
) (74
) (75
) (76
) (77
)8(80
) (81
) (82
) (83
) (84
) (85
) (86
) (87
) (88
)...
......
......
......
......
. . .
![Page 253: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/253.jpg)
Pascal’s Identity
The relationship between terms in row n − 1 and row n iscalled Pascal’s identityPascal’s identity states that(
n − 1m − 1
)+
(n − 1m
)=
(nm
)That is, the sum of terms in columns m − 1 and m of rown − 1 equals the term in column m row n
![Page 254: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/254.jpg)
Pascal’s Identity
The relationship between terms in row n − 1 and row n iscalled Pascal’s identity
Pascal’s identity states that(n − 1m − 1
)+
(n − 1m
)=
(nm
)That is, the sum of terms in columns m − 1 and m of rown − 1 equals the term in column m row n
![Page 255: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/255.jpg)
Pascal’s Identity
The relationship between terms in row n − 1 and row n iscalled Pascal’s identityPascal’s identity states that(
n − 1m − 1
)+
(n − 1m
)=
(nm
)
That is, the sum of terms in columns m − 1 and m of rown − 1 equals the term in column m row n
![Page 256: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/256.jpg)
Pascal’s Identity
The relationship between terms in row n − 1 and row n iscalled Pascal’s identityPascal’s identity states that(
n − 1m − 1
)+
(n − 1m
)=
(nm
)That is, the sum of terms in columns m − 1 and m of rown − 1 equals the term in column m row n
![Page 257: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/257.jpg)
Pascal’s Identity
An example of Pascal’s identity is(74
)=
7!4!3!
=7× 6× 53× 2× 1
= 35
and (75
)=
7!5!2!
=7× 62× 1
= 21
Therefore (85
)=
(74
)+
(75
)= 35 + 21
![Page 258: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/258.jpg)
Pascal’s Identity
An example of Pascal’s identity is(74
)=
7!4!3!
=7× 6× 53× 2× 1
= 35
and (75
)=
7!5!2!
=7× 62× 1
= 21
Therefore (85
)=
(74
)+
(75
)= 35 + 21
![Page 259: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/259.jpg)
Pascal’s Identity
Another example of Pascal’s identity is(128
)=
12!8!4!
= 495
and (129
)=
12!9!3!
= 660
Therefore (139
)= 495 + 660 = 1155
![Page 260: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/260.jpg)
Pascal’s Identity
Another example of Pascal’s identity is(128
)=
12!8!4!
= 495
and (129
)=
12!9!3!
= 660
Therefore (139
)= 495 + 660 = 1155
![Page 261: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/261.jpg)
Pascal’s Triangle Row Sum
Armed with Pascal’s identity, we can use induction to establishthe sum of values in row n is 2n
To sum the elements in row n we writen∑
k=0
(nk
)=
(n0
)+
(n1
)+
(n2
)+ · · ·+
(n
n − 1
)+
(nn
)Leave the first and last terms alone, but use Pascal’s identityon the middle terms(
n0
)+
[(n − 10
)+
(n − 11
)]+
[(n − 11
)+
(n − 12
)]+ · · ·+
[(n − 1n − 2
)+
(n − 1n − 1
)]+
(nn
)
![Page 262: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/262.jpg)
Pascal’s Triangle Row Sum
Armed with Pascal’s identity, we can use induction to establishthe sum of values in row n is 2n
To sum the elements in row n we writen∑
k=0
(nk
)=
(n0
)+
(n1
)+
(n2
)+ · · ·+
(n
n − 1
)+
(nn
)Leave the first and last terms alone, but use Pascal’s identityon the middle terms(
n0
)+
[(n − 10
)+
(n − 11
)]+
[(n − 11
)+
(n − 12
)]+ · · ·+
[(n − 1n − 2
)+
(n − 1n − 1
)]+
(nn
)
![Page 263: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/263.jpg)
Pascal’s Triangle Row Sum
Armed with Pascal’s identity, we can use induction to establishthe sum of values in row n is 2n
To sum the elements in row n we writen∑
k=0
(nk
)=
(n0
)+
(n1
)+
(n2
)+ · · ·+
(n
n − 1
)+
(nn
)
Leave the first and last terms alone, but use Pascal’s identityon the middle terms(
n0
)+
[(n − 10
)+
(n − 11
)]+
[(n − 11
)+
(n − 12
)]+ · · ·+
[(n − 1n − 2
)+
(n − 1n − 1
)]+
(nn
)
![Page 264: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/264.jpg)
Pascal’s Triangle Row Sum
Armed with Pascal’s identity, we can use induction to establishthe sum of values in row n is 2n
To sum the elements in row n we writen∑
k=0
(nk
)=
(n0
)+
(n1
)+
(n2
)+ · · ·+
(n
n − 1
)+
(nn
)Leave the first and last terms alone, but use Pascal’s identityon the middle terms(
n0
)+
[(n − 10
)+
(n − 11
)]+
[(n − 11
)+
(n − 12
)]+ · · ·+
[(n − 1n − 2
)+
(n − 1n − 1
)]+
(nn
)
![Page 265: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/265.jpg)
Pascal’s Triangle Row Sum
Collect the terms that appear twice(n0
)+
(n − 10
)+2(n − 11
)+· · ·+2
(n − 1n − 2
)+
(n − 1n − 1
)+
(nn
)Recognize that the first two and last two terms are equal to 1and so can be replaced as
2(n − 10
)+ 2(n − 11
)+ · · ·+ 2
(n − 1n − 2
)+ 2(n − 1n − 1
)Therefore, if the sum of terms in row n− 1 is 2n−1, the sum ofterms in row n is 2n
![Page 266: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/266.jpg)
Pascal’s Triangle Row Sum
Collect the terms that appear twice(n0
)+
(n − 10
)+2(n − 11
)+· · ·+2
(n − 1n − 2
)+
(n − 1n − 1
)+
(nn
)
Recognize that the first two and last two terms are equal to 1and so can be replaced as
2(n − 10
)+ 2(n − 11
)+ · · ·+ 2
(n − 1n − 2
)+ 2(n − 1n − 1
)Therefore, if the sum of terms in row n− 1 is 2n−1, the sum ofterms in row n is 2n
![Page 267: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/267.jpg)
Pascal’s Triangle Row Sum
Collect the terms that appear twice(n0
)+
(n − 10
)+2(n − 11
)+· · ·+2
(n − 1n − 2
)+
(n − 1n − 1
)+
(nn
)Recognize that the first two and last two terms are equal to 1and so can be replaced as
2(n − 10
)+ 2(n − 11
)+ · · ·+ 2
(n − 1n − 2
)+ 2(n − 1n − 1
)
Therefore, if the sum of terms in row n− 1 is 2n−1, the sum ofterms in row n is 2n
![Page 268: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/268.jpg)
Pascal’s Triangle Row Sum
Collect the terms that appear twice(n0
)+
(n − 10
)+2(n − 11
)+· · ·+2
(n − 1n − 2
)+
(n − 1n − 1
)+
(nn
)Recognize that the first two and last two terms are equal to 1and so can be replaced as
2(n − 10
)+ 2(n − 11
)+ · · ·+ 2
(n − 1n − 2
)+ 2(n − 1n − 1
)Therefore, if the sum of terms in row n− 1 is 2n−1, the sum ofterms in row n is 2n
![Page 269: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/269.jpg)
Epigraph
φ is an H of a lot cooler than π.
Stettner, in Dan Brown’s “The Da Vinci Code”
![Page 270: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/270.jpg)
Cassini’s Identity
Theorem (Cassini’s Identity)
Fn+1Fn−1 − F 2n = (−1)n, for n > 0
Interpret Cassini’s identity as stating the area of a rectanglewith sides Fn−1 and Fn+1 is plus or minus one the area of asquare with sides Fn
Arithmetically, multiply every other Fibonacci number andsubtract the square of the number in the middleThe result alternates between −1 and +1
![Page 271: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/271.jpg)
Cassini’s Identity
Theorem (Cassini’s Identity)
Fn+1Fn−1 − F 2n = (−1)n, for n > 0
Interpret Cassini’s identity as stating the area of a rectanglewith sides Fn−1 and Fn+1 is plus or minus one the area of asquare with sides Fn
Arithmetically, multiply every other Fibonacci number andsubtract the square of the number in the middleThe result alternates between −1 and +1
![Page 272: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/272.jpg)
Cassini’s Identity
Theorem (Cassini’s Identity)
Fn+1Fn−1 − F 2n = (−1)n, for n > 0
Interpret Cassini’s identity as stating the area of a rectanglewith sides Fn−1 and Fn+1 is plus or minus one the area of asquare with sides Fn
Arithmetically, multiply every other Fibonacci number andsubtract the square of the number in the middle
The result alternates between −1 and +1
![Page 273: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/273.jpg)
Cassini’s Identity
Theorem (Cassini’s Identity)
Fn+1Fn−1 − F 2n = (−1)n, for n > 0
Interpret Cassini’s identity as stating the area of a rectanglewith sides Fn−1 and Fn+1 is plus or minus one the area of asquare with sides Fn
Arithmetically, multiply every other Fibonacci number andsubtract the square of the number in the middleThe result alternates between −1 and +1
![Page 274: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/274.jpg)
Cassini’s Identity
n 0 1 2 3 4 5 6 7 8 9 . . .
Fn 0 1 1 2 3 5 8 13 21 34 . . .
0 · 1− 12 = −12 · 1− 12 = 1
3 · 1− 22 = −15 · 2− 32 = 1
8 · 3− 52 = −1
![Page 275: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/275.jpg)
Cassini’s Identity
n 0 1 2 3 4 5 6 7 8 9 . . .
Fn 0 1 1 2 3 5 8 13 21 34 . . .
0 · 1− 12 = −12 · 1− 12 = 1
3 · 1− 22 = −15 · 2− 32 = 1
8 · 3− 52 = −1
![Page 276: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/276.jpg)
Cassini’s Identity
n 0 1 2 3 4 5 6 7 8 9 . . .
Fn 0 1 1 2 3 5 8 13 21 34 . . .
0 · 1− 12
= −12 · 1− 12 = 1
3 · 1− 22 = −15 · 2− 32 = 1
8 · 3− 52 = −1
![Page 277: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/277.jpg)
Cassini’s Identity
n 0 1 2 3 4 5 6 7 8 9 . . .
Fn 0 1 1 2 3 5 8 13 21 34 . . .
0 · 1− 12 = −1
2 · 1− 12 = 1
3 · 1− 22 = −15 · 2− 32 = 1
8 · 3− 52 = −1
![Page 278: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/278.jpg)
Cassini’s Identity
n 0 1 2 3 4 5 6 7 8 9 . . .
Fn 0 1 1 2 3 5 8 13 21 34 . . .
0 · 1− 12 = −12 · 1− 12
= 1
3 · 1− 22 = −15 · 2− 32 = 1
8 · 3− 52 = −1
![Page 279: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/279.jpg)
Cassini’s Identity
n 0 1 2 3 4 5 6 7 8 9 . . .
Fn 0 1 1 2 3 5 8 13 21 34 . . .
0 · 1− 12 = −12 · 1− 12 = 1
3 · 1− 22 = −15 · 2− 32 = 1
8 · 3− 52 = −1
![Page 280: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/280.jpg)
Cassini’s Identity
n 0 1 2 3 4 5 6 7 8 9 . . .
Fn 0 1 1 2 3 5 8 13 21 34 . . .
0 · 1− 12 = −12 · 1− 12 = 1
3 · 1− 22
= −15 · 2− 32 = 1
8 · 3− 52 = −1
![Page 281: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/281.jpg)
Cassini’s Identity
n 0 1 2 3 4 5 6 7 8 9 . . .
Fn 0 1 1 2 3 5 8 13 21 34 . . .
0 · 1− 12 = −12 · 1− 12 = 1
3 · 1− 22 = −1
5 · 2− 32 = 1
8 · 3− 52 = −1
![Page 282: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/282.jpg)
Cassini’s Identity
n 0 1 2 3 4 5 6 7 8 9 . . .
Fn 0 1 1 2 3 5 8 13 21 34 . . .
0 · 1− 12 = −12 · 1− 12 = 1
3 · 1− 22 = −15 · 2− 32
= 1
8 · 3− 52 = −1
![Page 283: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/283.jpg)
Cassini’s Identity
n 0 1 2 3 4 5 6 7 8 9 . . .
Fn 0 1 1 2 3 5 8 13 21 34 . . .
0 · 1− 12 = −12 · 1− 12 = 1
3 · 1− 22 = −15 · 2− 32 = 1
8 · 3− 52 = −1
![Page 284: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/284.jpg)
Cassini’s Identity
n 0 1 2 3 4 5 6 7 8 9 . . .
Fn 0 1 1 2 3 5 8 13 21 34 . . .
0 · 1− 12 = −12 · 1− 12 = 1
3 · 1− 22 = −15 · 2− 32 = 1
8 · 3− 52
= −1
![Page 285: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/285.jpg)
Cassini’s Identity
n 0 1 2 3 4 5 6 7 8 9 . . .
Fn 0 1 1 2 3 5 8 13 21 34 . . .
0 · 1− 12 = −12 · 1− 12 = 1
3 · 1− 22 = −15 · 2− 32 = 1
8 · 3− 52 = −1
![Page 286: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/286.jpg)
Lewis Carroll’s Favorite Trick
Cassini’s identity is the basis of an absurdity attributed LewisCarroll
Cut the 8× 8 square along the lines indicated below andarrange the pieces into a 5× 13 rectangle to conclude that64 = 65.
![Page 287: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/287.jpg)
Lewis Carroll’s Favorite Trick
Cassini’s identity is the basis of an absurdity attributed LewisCarrollCut the 8× 8 square along the lines indicated below andarrange the pieces into a 5× 13 rectangle to conclude that64 = 65.
![Page 288: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/288.jpg)
Inductive Proof of Cassini’s Identity
Fn+1Fn−1 − F 2n = (−1)n, for n > 0
Replace Fn−1 by Fn+1 − Fn in the left-hand side of Cassini’sproposed identity to obtain
Fn+1Fn−1 − F 2n = Fn+1(Fn+1 − Fn)− F 2
n
= F 2n+1 − Fn+1Fn − F 2
n
= F 2n+1 − Fn(Fn+1 + Fn)
= F 2n+1 − FnFn+2
= −(FnFn+2 − F 2n+1)
= −(Fn+2Fn − F 2n+1)
![Page 289: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/289.jpg)
Inductive Proof of Cassini’s Identity
Fn+1Fn−1 − F 2n = (−1)n, for n > 0
Replace Fn−1 by Fn+1 − Fn in the left-hand side of Cassini’sproposed identity to obtain
Fn+1Fn−1 − F 2n = Fn+1(Fn+1 − Fn)− F 2
n
= F 2n+1 − Fn+1Fn − F 2
n
= F 2n+1 − Fn(Fn+1 + Fn)
= F 2n+1 − FnFn+2
= −(FnFn+2 − F 2n+1)
= −(Fn+2Fn − F 2n+1)
![Page 290: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/290.jpg)
Inductive Proof of Cassini’s Identity
Fn+1Fn−1 − F 2n = (−1)n, for n > 0
Replace Fn−1 by Fn+1 − Fn in the left-hand side of Cassini’sproposed identity to obtain
Fn+1Fn−1 − F 2n
= Fn+1(Fn+1 − Fn)− F 2n
= F 2n+1 − Fn+1Fn − F 2
n
= F 2n+1 − Fn(Fn+1 + Fn)
= F 2n+1 − FnFn+2
= −(FnFn+2 − F 2n+1)
= −(Fn+2Fn − F 2n+1)
![Page 291: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/291.jpg)
Inductive Proof of Cassini’s Identity
Fn+1Fn−1 − F 2n = (−1)n, for n > 0
Replace Fn−1 by Fn+1 − Fn in the left-hand side of Cassini’sproposed identity to obtain
Fn+1Fn−1 − F 2n = Fn+1(Fn+1 − Fn)− F 2
n
= F 2n+1 − Fn+1Fn − F 2
n
= F 2n+1 − Fn(Fn+1 + Fn)
= F 2n+1 − FnFn+2
= −(FnFn+2 − F 2n+1)
= −(Fn+2Fn − F 2n+1)
![Page 292: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/292.jpg)
Inductive Proof of Cassini’s Identity
Fn+1Fn−1 − F 2n = (−1)n, for n > 0
Replace Fn−1 by Fn+1 − Fn in the left-hand side of Cassini’sproposed identity to obtain
Fn+1Fn−1 − F 2n = Fn+1(Fn+1 − Fn)− F 2
n
= F 2n+1 − Fn+1Fn − F 2
n
= F 2n+1 − Fn(Fn+1 + Fn)
= F 2n+1 − FnFn+2
= −(FnFn+2 − F 2n+1)
= −(Fn+2Fn − F 2n+1)
![Page 293: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/293.jpg)
Inductive Proof of Cassini’s Identity
Fn+1Fn−1 − F 2n = (−1)n, for n > 0
Replace Fn−1 by Fn+1 − Fn in the left-hand side of Cassini’sproposed identity to obtain
Fn+1Fn−1 − F 2n = Fn+1(Fn+1 − Fn)− F 2
n
= F 2n+1 − Fn+1Fn − F 2
n
= F 2n+1 − Fn(Fn+1 + Fn)
= F 2n+1 − FnFn+2
= −(FnFn+2 − F 2n+1)
= −(Fn+2Fn − F 2n+1)
![Page 294: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/294.jpg)
Inductive Proof of Cassini’s Identity
Fn+1Fn−1 − F 2n = (−1)n, for n > 0
Replace Fn−1 by Fn+1 − Fn in the left-hand side of Cassini’sproposed identity to obtain
Fn+1Fn−1 − F 2n = Fn+1(Fn+1 − Fn)− F 2
n
= F 2n+1 − Fn+1Fn − F 2
n
= F 2n+1 − Fn(Fn+1 + Fn)
= F 2n+1 − FnFn+2
= −(FnFn+2 − F 2n+1)
= −(Fn+2Fn − F 2n+1)
![Page 295: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/295.jpg)
Inductive Proof of Cassini’s Identity
Fn+1Fn−1 − F 2n = (−1)n, for n > 0
Replace Fn−1 by Fn+1 − Fn in the left-hand side of Cassini’sproposed identity to obtain
Fn+1Fn−1 − F 2n = Fn+1(Fn+1 − Fn)− F 2
n
= F 2n+1 − Fn+1Fn − F 2
n
= F 2n+1 − Fn(Fn+1 + Fn)
= F 2n+1 − FnFn+2
= −(FnFn+2 − F 2n+1)
= −(Fn+2Fn − F 2n+1)
![Page 296: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/296.jpg)
Inductive Proof of Cassini’s Identity
Fn+1Fn−1 − F 2n = (−1)n, for n > 0
Replace Fn−1 by Fn+1 − Fn in the left-hand side of Cassini’sproposed identity to obtain
Fn+1Fn−1 − F 2n = Fn+1(Fn+1 − Fn)− F 2
n
= F 2n+1 − Fn+1Fn − F 2
n
= F 2n+1 − Fn(Fn+1 + Fn)
= F 2n+1 − FnFn+2
= −(FnFn+2 − F 2n+1)
= −(Fn+2Fn − F 2n+1)
![Page 297: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/297.jpg)
Inductive Proof of Cassini’s Identity
Fn+1Fn−1 − F 2n = (−1)n, for n > 0
Thus ifFn+1Fn−1 − F 2
n = (−1)n
ThenFn+2Fn − F 2
n+1 = (−1)n+1
Since F2F0 − F1 = 1 · 0− 1 = (−1)1, the basis for induction istrue, so Cassini’s identity holds for all n.
![Page 298: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/298.jpg)
Inductive Proof of Cassini’s Identity
Fn+1Fn−1 − F 2n = (−1)n, for n > 0
Thus ifFn+1Fn−1 − F 2
n = (−1)n
ThenFn+2Fn − F 2
n+1 = (−1)n+1
Since F2F0 − F1 = 1 · 0− 1 = (−1)1, the basis for induction istrue, so Cassini’s identity holds for all n.
![Page 299: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/299.jpg)
Inductive Proof of Cassini’s Identity
Fn+1Fn−1 − F 2n = (−1)n, for n > 0
Thus ifFn+1Fn−1 − F 2
n = (−1)n
ThenFn+2Fn − F 2
n+1 = (−1)n+1
Since F2F0 − F1 = 1 · 0− 1 = (−1)1, the basis for induction istrue, so Cassini’s identity holds for all n.
![Page 300: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/300.jpg)
Fundamental Theorem of Arithmetic
The fundamental theorem of arithmetic states that everypositive integer greater than 1 can be written as a (unique)product of primes
As a basis 2 is prime, hence a product of primesPretend that for some n ≥ 2, every integer k = 2, 3, . . . , ncan be written as the product of primesConsider n + 1
If n + 1 is prime, then it is the product of primesIf n + 1 is composite, then n + 1 = ab where 1 < a < n + 1and 1 < b < n + 1Therefore, by the inductive assumption, a and b are theproduct of primes, and so n + 1 is the product of primes
![Page 301: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/301.jpg)
Fundamental Theorem of Arithmetic
The fundamental theorem of arithmetic states that everypositive integer greater than 1 can be written as a (unique)product of primesAs a basis 2 is prime, hence a product of primes
Pretend that for some n ≥ 2, every integer k = 2, 3, . . . , ncan be written as the product of primesConsider n + 1
If n + 1 is prime, then it is the product of primesIf n + 1 is composite, then n + 1 = ab where 1 < a < n + 1and 1 < b < n + 1Therefore, by the inductive assumption, a and b are theproduct of primes, and so n + 1 is the product of primes
![Page 302: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/302.jpg)
Fundamental Theorem of Arithmetic
The fundamental theorem of arithmetic states that everypositive integer greater than 1 can be written as a (unique)product of primesAs a basis 2 is prime, hence a product of primesPretend that for some n ≥ 2, every integer k = 2, 3, . . . , ncan be written as the product of primes
Consider n + 1
If n + 1 is prime, then it is the product of primesIf n + 1 is composite, then n + 1 = ab where 1 < a < n + 1and 1 < b < n + 1Therefore, by the inductive assumption, a and b are theproduct of primes, and so n + 1 is the product of primes
![Page 303: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/303.jpg)
Fundamental Theorem of Arithmetic
The fundamental theorem of arithmetic states that everypositive integer greater than 1 can be written as a (unique)product of primesAs a basis 2 is prime, hence a product of primesPretend that for some n ≥ 2, every integer k = 2, 3, . . . , ncan be written as the product of primesConsider n + 1
If n + 1 is prime, then it is the product of primesIf n + 1 is composite, then n + 1 = ab where 1 < a < n + 1and 1 < b < n + 1Therefore, by the inductive assumption, a and b are theproduct of primes, and so n + 1 is the product of primes
![Page 304: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/304.jpg)
Fundamental Theorem of Arithmetic
The fundamental theorem of arithmetic states that everypositive integer greater than 1 can be written as a (unique)product of primesAs a basis 2 is prime, hence a product of primesPretend that for some n ≥ 2, every integer k = 2, 3, . . . , ncan be written as the product of primesConsider n + 1
If n + 1 is prime, then it is the product of primes
If n + 1 is composite, then n + 1 = ab where 1 < a < n + 1and 1 < b < n + 1Therefore, by the inductive assumption, a and b are theproduct of primes, and so n + 1 is the product of primes
![Page 305: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/305.jpg)
Fundamental Theorem of Arithmetic
The fundamental theorem of arithmetic states that everypositive integer greater than 1 can be written as a (unique)product of primesAs a basis 2 is prime, hence a product of primesPretend that for some n ≥ 2, every integer k = 2, 3, . . . , ncan be written as the product of primesConsider n + 1
If n + 1 is prime, then it is the product of primesIf n + 1 is composite, then n + 1 = ab where 1 < a < n + 1and 1 < b < n + 1
Therefore, by the inductive assumption, a and b are theproduct of primes, and so n + 1 is the product of primes
![Page 306: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/306.jpg)
Fundamental Theorem of Arithmetic
The fundamental theorem of arithmetic states that everypositive integer greater than 1 can be written as a (unique)product of primesAs a basis 2 is prime, hence a product of primesPretend that for some n ≥ 2, every integer k = 2, 3, . . . , ncan be written as the product of primesConsider n + 1
If n + 1 is prime, then it is the product of primesIf n + 1 is composite, then n + 1 = ab where 1 < a < n + 1and 1 < b < n + 1Therefore, by the inductive assumption, a and b are theproduct of primes, and so n + 1 is the product of primes
![Page 307: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/307.jpg)
The Fundamental Theorem of the Sum and DifferenceCalculus
Define the falling factorial power nm by
nm = n(n − 1) · · · (n −m + 1)
The following identity is true
mn−1∑k=0
km−1m =n−1∑k=0
k(k − 1) · · · (k −m + 2) = nm
As a basis for induction, when n = 0 the left-hand side equalsthe right-hand side
![Page 308: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/308.jpg)
The Fundamental Theorem of the Sum and DifferenceCalculus
Define the falling factorial power nm by
nm = n(n − 1) · · · (n −m + 1)
The following identity is true
mn−1∑k=0
km−1m =n−1∑k=0
k(k − 1) · · · (k −m + 2) = nm
As a basis for induction, when n = 0 the left-hand side equalsthe right-hand side
![Page 309: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/309.jpg)
The Fundamental Theorem of the Sum and DifferenceCalculus
Define the falling factorial power nm by
nm = n(n − 1) · · · (n −m + 1)
The following identity is true
mn−1∑k=0
km−1m =n−1∑k=0
k(k − 1) · · · (k −m + 2) = nm
As a basis for induction, when n = 0 the left-hand side equalsthe right-hand side
![Page 310: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/310.jpg)
The Fundamental Theorem of the Sum and DifferenceCalculus
Pretend
mn−1∑k=0
km−1 = nm
is true for some n ≥ 0
Then
mn∑
k=0
km−1 = mn−1∑k=0
km−1 + mnm−1
= nm + mnm−1
= [n(n − 1) · · · (n −m + 1)] + [m(n(n − 1) · · · (n −m + 2))]= n(n − 1) · · · (n −m + 2)[(n −m + 1) + m]
= (n + 1)n(n − 1) · · · (n −m + 2)= (n + 1)m
![Page 311: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/311.jpg)
The Fundamental Theorem of the Sum and DifferenceCalculus
Pretend
mn−1∑k=0
km−1 = nm
is true for some n ≥ 0Then
mn∑
k=0
km−1 = mn−1∑k=0
km−1 + mnm−1
= nm + mnm−1
= [n(n − 1) · · · (n −m + 1)] + [m(n(n − 1) · · · (n −m + 2))]= n(n − 1) · · · (n −m + 2)[(n −m + 1) + m]
= (n + 1)n(n − 1) · · · (n −m + 2)= (n + 1)m
![Page 312: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/312.jpg)
The Fundamental Theorem of the Sum and DifferenceCalculus
Pretend
mn−1∑k=0
km−1 = nm
is true for some n ≥ 0Then
mn∑
k=0
km−1 = mn−1∑k=0
km−1 + mnm−1
= nm + mnm−1
= [n(n − 1) · · · (n −m + 1)] + [m(n(n − 1) · · · (n −m + 2))]= n(n − 1) · · · (n −m + 2)[(n −m + 1) + m]
= (n + 1)n(n − 1) · · · (n −m + 2)= (n + 1)m
![Page 313: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/313.jpg)
The Fundamental Theorem of the Sum and DifferenceCalculus
Pretend
mn−1∑k=0
km−1 = nm
is true for some n ≥ 0Then
mn∑
k=0
km−1
= mn−1∑k=0
km−1 + mnm−1
= nm + mnm−1
= [n(n − 1) · · · (n −m + 1)] + [m(n(n − 1) · · · (n −m + 2))]= n(n − 1) · · · (n −m + 2)[(n −m + 1) + m]
= (n + 1)n(n − 1) · · · (n −m + 2)= (n + 1)m
![Page 314: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/314.jpg)
The Fundamental Theorem of the Sum and DifferenceCalculus
Pretend
mn−1∑k=0
km−1 = nm
is true for some n ≥ 0Then
mn∑
k=0
km−1 = mn−1∑k=0
km−1 + mnm−1
= nm + mnm−1
= [n(n − 1) · · · (n −m + 1)] + [m(n(n − 1) · · · (n −m + 2))]= n(n − 1) · · · (n −m + 2)[(n −m + 1) + m]
= (n + 1)n(n − 1) · · · (n −m + 2)= (n + 1)m
![Page 315: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/315.jpg)
The Fundamental Theorem of the Sum and DifferenceCalculus
Pretend
mn−1∑k=0
km−1 = nm
is true for some n ≥ 0Then
mn∑
k=0
km−1 = mn−1∑k=0
km−1 + mnm−1
= nm + mnm−1
= [n(n − 1) · · · (n −m + 1)] + [m(n(n − 1) · · · (n −m + 2))]= n(n − 1) · · · (n −m + 2)[(n −m + 1) + m]
= (n + 1)n(n − 1) · · · (n −m + 2)= (n + 1)m
![Page 316: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/316.jpg)
The Fundamental Theorem of the Sum and DifferenceCalculus
Pretend
mn−1∑k=0
km−1 = nm
is true for some n ≥ 0Then
mn∑
k=0
km−1 = mn−1∑k=0
km−1 + mnm−1
= nm + mnm−1
= [n(n − 1) · · · (n −m + 1)] + [m(n(n − 1) · · · (n −m + 2))]
= n(n − 1) · · · (n −m + 2)[(n −m + 1) + m]
= (n + 1)n(n − 1) · · · (n −m + 2)= (n + 1)m
![Page 317: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/317.jpg)
The Fundamental Theorem of the Sum and DifferenceCalculus
Pretend
mn−1∑k=0
km−1 = nm
is true for some n ≥ 0Then
mn∑
k=0
km−1 = mn−1∑k=0
km−1 + mnm−1
= nm + mnm−1
= [n(n − 1) · · · (n −m + 1)] + [m(n(n − 1) · · · (n −m + 2))]= n(n − 1) · · · (n −m + 2)[(n −m + 1) + m]
= (n + 1)n(n − 1) · · · (n −m + 2)= (n + 1)m
![Page 318: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/318.jpg)
The Fundamental Theorem of the Sum and DifferenceCalculus
Pretend
mn−1∑k=0
km−1 = nm
is true for some n ≥ 0Then
mn∑
k=0
km−1 = mn−1∑k=0
km−1 + mnm−1
= nm + mnm−1
= [n(n − 1) · · · (n −m + 1)] + [m(n(n − 1) · · · (n −m + 2))]= n(n − 1) · · · (n −m + 2)[(n −m + 1) + m]
= (n + 1)n(n − 1) · · · (n −m + 2)
= (n + 1)m
![Page 319: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/319.jpg)
The Fundamental Theorem of the Sum and DifferenceCalculus
Pretend
mn−1∑k=0
km−1 = nm
is true for some n ≥ 0Then
mn∑
k=0
km−1 = mn−1∑k=0
km−1 + mnm−1
= nm + mnm−1
= [n(n − 1) · · · (n −m + 1)] + [m(n(n − 1) · · · (n −m + 2))]= n(n − 1) · · · (n −m + 2)[(n −m + 1) + m]
= (n + 1)n(n − 1) · · · (n −m + 2)= (n + 1)m
![Page 320: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/320.jpg)
The Fundamental Theorem of the Sum and DifferenceCalculus
The fundamental theorem of the sum and difference calculus is
mn−1∑k=0
km−1 = nm
Notice the analogy with the fundamental theorem of calculus
m∫
xm−1dx = xm
![Page 321: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/321.jpg)
The Fundamental Theorem of the Sum and DifferenceCalculus
The fundamental theorem of the sum and difference calculus is
mn−1∑k=0
km−1 = nm
Notice the analogy with the fundamental theorem of calculus
m∫
xm−1dx = xm
![Page 322: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/322.jpg)
Exponential Beat Powers
The exponent 2n grows faster than the polynomial n2
Prove n2 ≤ 2n for all n > 3
For n = 4 we have 42 = 1624
Pretend n2 ≤ 2n for some n ≥ 4Notice that
(n + 1)2 ≤ 2n2, when n > 3
since
0 ≤ 2n2 − (n + 1)2 = n2 − 2n − 1 = (n − 1)2 − 2
Therefore
(n + 1)2 = 2n2
≤ 2 · 2n
= 2n+1
![Page 323: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/323.jpg)
Exponential Beat Powers
The exponent 2n grows faster than the polynomial n2
Prove n2 ≤ 2n for all n > 3
For n = 4 we have 42 = 1624
Pretend n2 ≤ 2n for some n ≥ 4Notice that
(n + 1)2 ≤ 2n2, when n > 3
since
0 ≤ 2n2 − (n + 1)2 = n2 − 2n − 1 = (n − 1)2 − 2
Therefore
(n + 1)2 = 2n2
≤ 2 · 2n
= 2n+1
![Page 324: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/324.jpg)
Exponential Beat Powers
The exponent 2n grows faster than the polynomial n2
Prove n2 ≤ 2n for all n > 3For n = 4 we have 42 = 1624
Pretend n2 ≤ 2n for some n ≥ 4Notice that
(n + 1)2 ≤ 2n2, when n > 3
since
0 ≤ 2n2 − (n + 1)2 = n2 − 2n − 1 = (n − 1)2 − 2
Therefore
(n + 1)2 = 2n2
≤ 2 · 2n
= 2n+1
![Page 325: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/325.jpg)
Exponential Beat Powers
The exponent 2n grows faster than the polynomial n2
Prove n2 ≤ 2n for all n > 3For n = 4 we have 42 = 1624
Pretend n2 ≤ 2n for some n ≥ 4
Notice that(n + 1)2 ≤ 2n2, when n > 3
since
0 ≤ 2n2 − (n + 1)2 = n2 − 2n − 1 = (n − 1)2 − 2
Therefore
(n + 1)2 = 2n2
≤ 2 · 2n
= 2n+1
![Page 326: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/326.jpg)
Exponential Beat Powers
The exponent 2n grows faster than the polynomial n2
Prove n2 ≤ 2n for all n > 3For n = 4 we have 42 = 1624
Pretend n2 ≤ 2n for some n ≥ 4Notice that
(n + 1)2 ≤ 2n2, when n > 3
since
0 ≤ 2n2 − (n + 1)2 = n2 − 2n − 1 = (n − 1)2 − 2
Therefore
(n + 1)2 = 2n2
≤ 2 · 2n
= 2n+1
![Page 327: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/327.jpg)
Exponential Beat Powers
The exponent 2n grows faster than the polynomial n2
Prove n2 ≤ 2n for all n > 3For n = 4 we have 42 = 1624
Pretend n2 ≤ 2n for some n ≥ 4Notice that
(n + 1)2 ≤ 2n2, when n > 3
since
0 ≤ 2n2 − (n + 1)2 = n2 − 2n − 1 = (n − 1)2 − 2
Therefore
(n + 1)2 = 2n2
≤ 2 · 2n
= 2n+1
![Page 328: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/328.jpg)
Exponential Beat Powers
The exponent 2n grows faster than the polynomial n2
Prove n2 ≤ 2n for all n > 3For n = 4 we have 42 = 1624
Pretend n2 ≤ 2n for some n ≥ 4Notice that
(n + 1)2 ≤ 2n2, when n > 3
since
0 ≤ 2n2 − (n + 1)2 = n2 − 2n − 1 = (n − 1)2 − 2
Therefore
(n + 1)2 = 2n2
≤ 2 · 2n
= 2n+1
![Page 329: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/329.jpg)
Exponential Beat Powers
The exponent 2n grows faster than the polynomial n2
Prove n2 ≤ 2n for all n > 3For n = 4 we have 42 = 1624
Pretend n2 ≤ 2n for some n ≥ 4Notice that
(n + 1)2 ≤ 2n2, when n > 3
since
0 ≤ 2n2 − (n + 1)2 = n2 − 2n − 1 = (n − 1)2 − 2
Therefore
(n + 1)2
= 2n2
≤ 2 · 2n
= 2n+1
![Page 330: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/330.jpg)
Exponential Beat Powers
The exponent 2n grows faster than the polynomial n2
Prove n2 ≤ 2n for all n > 3For n = 4 we have 42 = 1624
Pretend n2 ≤ 2n for some n ≥ 4Notice that
(n + 1)2 ≤ 2n2, when n > 3
since
0 ≤ 2n2 − (n + 1)2 = n2 − 2n − 1 = (n − 1)2 − 2
Therefore
(n + 1)2 = 2n2
≤ 2 · 2n
= 2n+1
![Page 331: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/331.jpg)
Exponential Beat Powers
The exponent 2n grows faster than the polynomial n2
Prove n2 ≤ 2n for all n > 3For n = 4 we have 42 = 1624
Pretend n2 ≤ 2n for some n ≥ 4Notice that
(n + 1)2 ≤ 2n2, when n > 3
since
0 ≤ 2n2 − (n + 1)2 = n2 − 2n − 1 = (n − 1)2 − 2
Therefore
(n + 1)2 = 2n2
≤ 2 · 2n
= 2n+1
![Page 332: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/332.jpg)
Exponential Beat Powers
The exponent 2n grows faster than the polynomial n2
Prove n2 ≤ 2n for all n > 3For n = 4 we have 42 = 1624
Pretend n2 ≤ 2n for some n ≥ 4Notice that
(n + 1)2 ≤ 2n2, when n > 3
since
0 ≤ 2n2 − (n + 1)2 = n2 − 2n − 1 = (n − 1)2 − 2
Therefore
(n + 1)2 = 2n2
≤ 2 · 2n
= 2n+1
![Page 333: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/333.jpg)
Induction Over Products
Define the product notation∏
ak by
n−1∏k=0
ak = a0a1 · · · an−1
Prove by induction that
n∏k=2
(1− 1
k2
)=
n + 12n
for n ≥ 2
For a basis, let n = 2, and find
2∏k=2
(1− 1
k2
)= 1− 1
22 =2 + 12 · 2
![Page 334: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/334.jpg)
Induction Over Products
Define the product notation∏
ak by
n−1∏k=0
ak = a0a1 · · · an−1
Prove by induction that
n∏k=2
(1− 1
k2
)=
n + 12n
for n ≥ 2
For a basis, let n = 2, and find
2∏k=2
(1− 1
k2
)= 1− 1
22 =2 + 12 · 2
![Page 335: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/335.jpg)
Induction Over Products
Define the product notation∏
ak by
n−1∏k=0
ak = a0a1 · · · an−1
Prove by induction that
n∏k=2
(1− 1
k2
)=
n + 12n
for n ≥ 2
For a basis, let n = 2, and find
2∏k=2
(1− 1
k2
)= 1− 1
22 =2 + 12 · 2
![Page 336: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/336.jpg)
Induction Over Products
Pretend thatn∏
k=2
(1− 1
k2
)=
n + 12n
for n ≥ 2
for some n ≥ 2
Then notice that
n+1∏k=2
(1− 1
k2
)=
[n∏
k=2
(1− 1
k2
)](1− 1
(n + 1)2
)=
[n + 12n
](1− 1
(n + 1)2
)=
12n
((n + 1)2 − 1
(n + 1)
)=
12n
(n2 + 2n(n + 1)
)=
n + 22(n + 1)
![Page 337: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/337.jpg)
Induction Over Products
Pretend thatn∏
k=2
(1− 1
k2
)=
n + 12n
for n ≥ 2
for some n ≥ 2Then notice that
n+1∏k=2
(1− 1
k2
)=
[n∏
k=2
(1− 1
k2
)](1− 1
(n + 1)2
)=
[n + 12n
](1− 1
(n + 1)2
)=
12n
((n + 1)2 − 1
(n + 1)
)=
12n
(n2 + 2n(n + 1)
)=
n + 22(n + 1)
![Page 338: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/338.jpg)
Induction Over Products
Pretend thatn∏
k=2
(1− 1
k2
)=
n + 12n
for n ≥ 2
for some n ≥ 2Then notice that
n+1∏k=2
(1− 1
k2
)=
[n∏
k=2
(1− 1
k2
)](1− 1
(n + 1)2
)=
[n + 12n
](1− 1
(n + 1)2
)=
12n
((n + 1)2 − 1
(n + 1)
)=
12n
(n2 + 2n(n + 1)
)=
n + 22(n + 1)
![Page 339: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/339.jpg)
Induction Over Products
Pretend thatn∏
k=2
(1− 1
k2
)=
n + 12n
for n ≥ 2
for some n ≥ 2Then notice that
n+1∏k=2
(1− 1
k2
)
=
[n∏
k=2
(1− 1
k2
)](1− 1
(n + 1)2
)=
[n + 12n
](1− 1
(n + 1)2
)=
12n
((n + 1)2 − 1
(n + 1)
)=
12n
(n2 + 2n(n + 1)
)=
n + 22(n + 1)
![Page 340: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/340.jpg)
Induction Over Products
Pretend thatn∏
k=2
(1− 1
k2
)=
n + 12n
for n ≥ 2
for some n ≥ 2Then notice that
n+1∏k=2
(1− 1
k2
)=
[n∏
k=2
(1− 1
k2
)](1− 1
(n + 1)2
)
=
[n + 12n
](1− 1
(n + 1)2
)=
12n
((n + 1)2 − 1
(n + 1)
)=
12n
(n2 + 2n(n + 1)
)=
n + 22(n + 1)
![Page 341: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/341.jpg)
Induction Over Products
Pretend thatn∏
k=2
(1− 1
k2
)=
n + 12n
for n ≥ 2
for some n ≥ 2Then notice that
n+1∏k=2
(1− 1
k2
)=
[n∏
k=2
(1− 1
k2
)](1− 1
(n + 1)2
)=
[n + 12n
](1− 1
(n + 1)2
)
=12n
((n + 1)2 − 1
(n + 1)
)=
12n
(n2 + 2n(n + 1)
)=
n + 22(n + 1)
![Page 342: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/342.jpg)
Induction Over Products
Pretend thatn∏
k=2
(1− 1
k2
)=
n + 12n
for n ≥ 2
for some n ≥ 2Then notice that
n+1∏k=2
(1− 1
k2
)=
[n∏
k=2
(1− 1
k2
)](1− 1
(n + 1)2
)=
[n + 12n
](1− 1
(n + 1)2
)=
12n
((n + 1)2 − 1
(n + 1)
)
=12n
(n2 + 2n(n + 1)
)=
n + 22(n + 1)
![Page 343: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/343.jpg)
Induction Over Products
Pretend thatn∏
k=2
(1− 1
k2
)=
n + 12n
for n ≥ 2
for some n ≥ 2Then notice that
n+1∏k=2
(1− 1
k2
)=
[n∏
k=2
(1− 1
k2
)](1− 1
(n + 1)2
)=
[n + 12n
](1− 1
(n + 1)2
)=
12n
((n + 1)2 − 1
(n + 1)
)=
12n
(n2 + 2n(n + 1)
)
=n + 2
2(n + 1)
![Page 344: Discrete Mathematics-Mathematical Induction](https://reader036.vdocuments.mx/reader036/viewer/2022081800/5468fd4db4af9fba2b8b45ca/html5/thumbnails/344.jpg)
Induction Over Products
Pretend thatn∏
k=2
(1− 1
k2
)=
n + 12n
for n ≥ 2
for some n ≥ 2Then notice that
n+1∏k=2
(1− 1
k2
)=
[n∏
k=2
(1− 1
k2
)](1− 1
(n + 1)2
)=
[n + 12n
](1− 1
(n + 1)2
)=
12n
((n + 1)2 − 1
(n + 1)
)=
12n
(n2 + 2n(n + 1)
)=
n + 22(n + 1)