mathematics. principle of mathematical induction session
TRANSCRIPT
Session Objective
1. Introduction
2. Steps involved in the use of mathematical induction
3. Principle of mathematical induction.
Statement
Statement:- A sentence which can be judged as true or false.
Example:1. ‘2 is only even prime number’ 2. ‘Bagdad is capital of Iraq’ 3. ‘2n+5 is always divisible by 5 for all nN
Mathematical statement:
Example 1 and 3.
Induction
Induction :It’s a process Particular General
Example: Statement- ’2n+1’ is odd number.
n=1 2.1+1=3 is odd. True
n=2 2.2+1=5 is odd. True
n=3 2.3+1=7 is odd. True
Observation tentative conclusion (‘2n+1 is odd’)
let its true for n=m. i.e 2m+1 is odd.
Induction
Steps Involved: 1. Verification2. Induction3. Generalization.
Important: Process of Mathematical Induction (PMI) is applicable for natural numbers.
Usage: 1. to prove mathematical formula
2. to check divisibility of a expression by a number
Ex:1.3+2.32+3.33+......+n.3n= n 1(2n 1)3 3
4
Ex: Prove n3+5n is divisible by ‘3’.
Algorithm
Let P(n) be the given statement.
Step 1: Prove P(1) is true Verification
Step 2: Assume P(n) is true for some n=mN i.e. P(m) is true.
Step 3: Using above assumption prove P(m+1) is true. i.e P(m) P (m+1)
Step 4: Above steps lead us to generalize the fact P(n) is true for all n N.
Illustrative Example
Principle of mathematical induction is applicable to
(a) set of integers
(b) set of real numbers
(c) set of positive integers
(d) None of these
Solution : (c)
Principle of mathematical induction is applicable to natural numbers or set of positive integers only.
Illustrative Example
Show by PMI that 1.3+2.32+3.33+......+n.3n=
n 1(2n 1)3 3
4Solution:
Step 1. for n=1, p(1)=1. 3=3 (LHS)
n 1 2(2n 1)3 3 (2 1 1)3 3R.H.S 3
4 4
L.H.S=R.H.S P(1) is true
Step2. Assume that P(m) is true
m 12 m (2m 1)3 3
P(m) : 1 .3 + 2 . 3 + .... + m .3 =4
Solution Continued
2 m
m 1
P(m) : 1.3 + 2.3 + .... + m.3
(2m 1)3 3=
4
Step3: To prove P(m + 1) holds true Adding. (m + 1).3m+1 to both sides
m 1 m 12 m m 1 (2m 1)3 3 4(m 1)3
1.3 2.3 ... m.3 m 1 .34
m 1 m 1(2m 1) 3 4(m 1)3 3
4
m 13 (2m 1 4m 4) 3
4
m 1(6m 3)3 3
4
m 1 m 23(2m 1) 3 3 (2m 1)3 3
4 4P(m) P (m+1)
Solution Continued
Step4. As P(m) P (m+1) P(n) is true for all n N
1.3+2.32+3.33+......+n.3n= n 1(2n 1)3 3
4
(Proved)
2 m m 1
m 2
1.3 2.3 ... m.3 m 1 .3
(2m 1)3 3
4
(m 1) 1{2(m 1) 1}3 3
4
Illustrative Example
Prove n3+5n is divisible by ‘3’ for n N (By PMI or Otherwise)
Solution: P(n) : ‘n3+5n is divisible by 3’
Step1: P(1) = ‘6 divisible by 3’ which is true
Step2: For some n=m, P(m) holds truei.e. m3+5m=3k, k N
step3: To prove P(m+1) holds true, we have to prove that (m+1)3+5(m+1) is divisible by 3.
= 3k´
(m+1)3+5(m+1)=(m3+5m)+(3m2+3m+6)
= 3k+3(m2+m+2)
( m2 + m + 2 I )
Solution Continued
(m+1)3+5(m+1)=3k’
P(m+1) is divisible by 3
P(m) P (m+1)
Step4: P(n) is true for all n N
n3+5n is divisible by ‘3’ for n N
Class Exercise - 6
Prove that mathematical induction that 72n + 3n – 1(23n – 3) is divisible by 25, .n N
Solution :
Let P(n) : 72n + (23n – 3)3n – 1 is divisible by 25.
Step I: n = 1
P(1) = 72 + (23 – 3)31 – 1
= 72 + 20 · 30
= 49 + 1 = 50 As P(1) is 50 which is divisible by 25, hence P(1) is true.
Solution Continued
Step II: Assuming P(m) is divisible by 25,
P(m) = 72m + (23m – 3)3m – 1 = 25(K) ... (i)
(K is a positive integer.)
Now P(m + 1) = 72(m + 1) + (23(m + 1) – 3)3m + 1 – 1
= 72m + 2 + (23m + 3 – 3)3m + 1 – 1
= 49 × 72m + 8(23m – 3)3m – 1 × 3
= 49 × 72m + 24(23m – 3)3m – 1
With the help of equation (i), we can write the above expression as
Solution Continued
3m 3 m 1 3m 3 m 149 25K 2 3 24 2 3
3m 3 m 1 3m 3 m 149 25K 49 2 3 24 2 3
3m 3 m 149 25K 2 3 49 24 3m 3 m 149 25K 25 2 3
3m 3 m 125 49k 2 3
3m 3 m 1P m 1 25 49k 2 3
Now from the above equation, we can conclude that P(m + 1) is divisible by 25.
Hence, P(n) is divisible by 25 for all natural numbers.
Alternative Solution
Alternative Method: without PMI
n3+5n = n(n2+5)
=n(n2 -1+6)
=n(n2-1)+6n
=n(n-1)(n+1)+6n
Product of three consecutive numbers
Illustrative Example
P(n) is the statement ‘n2 – n + 41 is prime’Verify it.
Solution:
For n = 1 P(1) = ‘41 is a prime’ True.
For n = 2 P(2) = ‘43 is a prime’ True.
But for n = 41 P(41) = ‘412 is a prime’ False.
False Statement
Class Exercise -3Prove by PMI that
1.2.3. + 2.3.4 + 3.4.5 + ... +
n(n + 1) (n + 2) = n(n 1) n 2 n 3
4
Solution:
step1. P(1): LHS=1.2.3=6
1.(1 1).(1 2).(1 3)R.H.S 6
4 L.H.S=R.H.S
Step2. Assume P(m) is true
m(m 1) m 2 m 31.2.3.+2.3.4+..+m(m+1)(m+ 2) =
4
Solution Continued
P(m): 1.2.3.+2.3.4+..+m(m+1)(m+ 2)
m(m 1) m 2 m 3=
4
Adding (m+1)(m+2)(m+3)
1.2.3.+2.3.4+...+m(m+1)(m+2)+(m+1)(m+2)(m+3)=
m(m 1) m 2 m 3(m 1) m 2 m 3
4
m(m 1) m 2 m 3 4(m 1) m 2 m 3
4
(m 1) m 2 m 3 (m 4)
4
Solution Continued
Step4. As P(m) P (m+1) P(n) is true for all n N
1.2.3.+2.3.4+...+(m+1)(m+2)(m+3)
(m 1) m 2 m 3 (m 4)
4
1.2.3.+2.3.4+3.4.5+...+n(n+1)(n+2)=
n(n 1) n 2 n 3
4
Class Exercise -4
If a+b=c+d and a2+b2=c2+d2 , then show by mathematical induction, an+bn = cn+dn
Solution: Let P(n) : an + bn = cn + dn
n = 2, P(2) : a2+b2 = c2+d2
For n = 1, P(1) : a+b=c+d
P(1) and P(2) hold true.
Assume P(m) and P(m + 1) hold true
am+bm = cm+dm
am+1+bm+1= cm+1+dm+1
Solution Continued
a+b = c+d; a2+b2=c2+d2
am+bm = cm+dm; am+1+bm+1= cm+1+dm+1
P(m+2) : am+2+bm+2
= (a+b)(am+1+bm+1)–ab(am+bm)
= (c+d)(cm+1+dm+1)–cd(cm+dm)
= cm + 2 + dm + 2
P(m+2) holds true.
an+bn = cn+dn holds true for n N
Class Exercise - 7
nn 1
n
UsingPMI prove that
sin2cos cos2 cos 4 ... cos2
2 sin
Solution:
nn 1
n
sin2Let P(n) : cos cos2 cos 4 ... cos2
2 sin
For n = 1, LHS = Cos
sin2R.H.S cos
2sin
Assume P(m) holds true
mm 1
m
sin2cos .cos2 .cos 4 ... cos2
2 sin
Solution Continued
mm 1
m
sin2cos .cos2 .cos 4 ... cos2
2 sin
multiplying both sides by Cos2m m 1 mcos cos2 cos 4 ...cos2 cos2
m m
m
sin2 .cos2
2 sin
2m m
m 1
2 sin cos2
2 sin
m 1
m 1
sin2
2 sin :P(m+1) holds true
As P(m) P(m+1) P(n) is true for all n N
nn 1
n
sin2cos .cos2 .cos 4 ... cos2
2 sin
Class Exercise - 8
21Prove that 1 + 2 + 3 + ... + n< (2n + 1)
8
Solution:
For n = 1, LHS = 1;RHS =9/8 LHS < RHS
Let assume P(m) in true
211 + 2 + 3 + ... + m< (2m + 1)
8
211 + 2 + 3 + ... + m+(m+1)< (2m + 1) (m 1)
8
24m 4m 1
m 18
Solution Continued
24m 4m 11 + 2 + 3 + ... + m+(m+1)< m 1
8
24m 12m 9
8
211 + 2 + 3 + ... + m+(m+1)< (2m 3)
8
211 + 2 + 3 + ... + m+(m+1)< {2(m 1) 1}
8
P (m+1) holds true
P(n) holds true n N
Class Exercise -9
n 1
Prove by PMI that
7 + 77 + 777 + ... + 777... 7 (n times)
7 = 10 9n 10
81
Solution:
For n = 1, LHS =7
RHS = 7
LHS = RHS
2710 9 10
81
Let P(n) holds true for n = m
P(m):7+77+777+...+77...7(m times) m 1710 9m 10
81
Solution Continued
7+77+777+...+77...7(m times)
m 1710 9m 10
81
Adding77...7(m+1)times to both sides
7+77+...+77...7(m times)+77...7(m + 1) times
m 1710 9m 10 + 777 ... 7 (m + 1) times
81
m 17 7= 10 9m 10 99...9(m 1)times
81 9
m 1 m 17 710 9m 10 10 1
81 9
Solution Continued
7+77+...+77...7(m + 1) times
m 1 m 17 710 9m 10 10 1
81 9
m 1 m 17
10 9m 10 9.10 981
m 17
10.10 9 m 1 1081
m 27
10 9 m 1 1081
P(m + 1) holds true P(n) is true
7+77+777+...+77...7(n times) n 1710 9m 10 n N
81
Class Exercise -10
2
1 1 1 1 11 ... 2 n 1; n N
4 9 16 nn
Prove that
Solution :- 2
1 1 1 1Let P(n) : 1 ... 2 n 1
4 9 nn
For n = 2, 1 5
LHS 14 4
1 3 6
RHS 22 2 4
LHS RHS
P(n) is true for n 2
2
1 1 1 1Let P(m) : 1 ... 2 for m 1....(1)
4 9 mm
2
1Now Adding to both side in equ. (1)
m 1
Solution Continued
2 2
1 1 1 1We get 1 ...
4 9 m m 1
2
1 12
m m 1
2
1 12
m m 1
2
2
m 2m 1 m2
2 m 1
2
m(m 1) 12
m m 1
2
1 1 12 2
m 1 m 1m m 1
2
10 as m N
m m 1 P(m + 1) holds true.
P(n) holds true , n > 1. n N