chapter 14 14...however • consider (a+b)5 • the first coefficient is 1 • the second...
TRANSCRIPT
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11
Chapter 14
Additional Topics in Algebra
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Chapter 14.1
Mathematical Induction
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Motivation
• Consider 1 + 3 + 5 + …(2n-1)
• 1 + 3 + 5 + 7 + 9 = 25 = 52
• 1 + 3 + 5 + 7 + 9 + 11 = 36 = 62
• Is 1 + 3 + 5 + …..+ (2n-1) = n2?
We will try to prove this by induction:
IF P1 is true AND
For all k, if Pk is true then Pk+1 is true
THEN Pn is true for all n
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Example
• 1 + 3 + 5 + …(2n-1) = n2
• Let n = 1: (2n-1) = 1= n2
• Assume 1 + 3 + …(2n-1) = n2
• Show 1 + 3 + … (2n-1) + (2(n+1)-1) = (n+1) 2
n2+ 2(n + 1) =n2 + 2n + 2 = (n+1) 2
• Expression remains true
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Example
• Show 2 + 4 + 6 +…+ 2n = n(n+1)
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Solution
• Show 2 + 4 + 6 +…+ 2n = n(n+1)
• n=1: 2n = 2, n(n+1) = 1(2) = 2, holds
• Given 2 + 4 + …2n = n (n+1) show
2 + 4 + …2(n+1) = (n+1)(n+1+1) = (n+1)(n+2)
• 2 + 4 + …+ 2n + 2(n+1) = n(n+1) + 2(n+1)
= n2 + n + 2n + 2 = n2 + 3n + 2 = (n+1)(n+2); works
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Example
• Show 23 + 43 + 63 + (2n) 3 = 2n2(n+1) 2
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Solution
• Show 23 + 43 + 63 + (2n) 3 = 2n2(n+1) 2
• n=1: (2) 3 = 2(1) 3(1+1) 2 or 8 = 2(4) = 8
• n+1:
• Add term to left side (2(n+1)) 3 gives 2 + 4 ..+ (2n) 3 +(2(n+1)) 3
• Add to right side; 2n 2(n+1) 2 + [2(n + 1)]3 =
2n2 (n+1)2 + 8(n+ 1) 3 = 2(n+1) 2 [ n 2 + 4(n+1)]= 2(n+1)2 (n+2) 2
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Example
• Prove 5 + 9 + 13 +…(4n+1) = n(2n+3)
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Solution
• Prove 5 + 9 + 13 +…(4n+1) = n(2n+3)
• n=1: 4(1)+1 = 5 = 1(2(1)+3) = 5
• Given works for n: 5 + 9 + … = n(2n+3), look at n+1
• 4(n+1) + 1 + n(2n+3) = 4n + 4 + 1 + 2n2 +3n
= 2n2 +7n + 5
• Right side = (n+1)(2(n+1) + 3) = (n+1)(2n + 2 + 3) =
(n+1)(2n + 5) = 2n2 + 7n + 5 works
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Example
• Prove 1x2 + 3x4 + 5x6 + (2n-1)(2n) = n(n+1)(4n-1)/3
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Solution
• Prove 1x2 + 3x4 + 5x6 + (2n-1)(2n) = n(n+1)(4n-1)/3
• n=1: (2-1)(2) = 2 = 1(1+1)(4-1)/3 = 2(3)/3 = 2 works
• For n: sum = n(n+1)(4n-1)/3
add n+1 term: n(n+1)(4n-1)/3 + 2(n+1)(2(n+1)-1)
=(n+1)[n(4n-1) + 2·3(2(n+1)-1)]/3
= (n+1)[4n2 – n + 12n + 12 – 6]/3 = (n+1)[4n2 + 11n +6]/3
• Sum for n+1 is (n+1)(n+2)(4(n+1)-1)/3 =
(n+1)(n+2)(4n + 4 – 1)/3 = (n+1)(n+2)(4n + 3)/3
= (n+1)(4n2 + 11n + 6)/3 works
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Chapter 14.2
The Binomial Theorem
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Binomial Theorem
• The value of (a+b)n can
• (a+b) = a + b
• (a+b) 2 = a2 + 2b + b2
• (a+b) 3 = a3 + 3a2b + 3ab2 + b3
• (a+b) 4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
• Note the pattern of the coefficients
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Pascal’s Triangle
• You can always find the coefficients by writing this triangle
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However
• Consider (a+b)5
• The first coefficient is 1
• The second coefficient is (the first coefficient:1) (order of a in
first: 6)/(number of first term: 1) = 1x6/1 = 6
• The third coefficient is (the 2nd coefficient: 6)
(order of a in 2nd term: 5) /(number of 2nd term: 2) = 6x5/2 = 15
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Factorial
• n! = 1 x 2 x 3 x 4 x …n
• 1! = 1
• 0! = 1 (by definition)
• Find (n+1)!/(n-1)!
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Example
• Find (n+1)!/(n-1)! = n(n+1)= n2 + n
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The Binomial Coefficient
• �� =
�!�! ��� ! for n ≥ k
• 53 =
!�!�!
·�·�·�·�(�·�·�)(�·�) =
·��·� = 10
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The Binomial Theorem
• (a+b)n =�0 an +
�1 an-1b +
�2 an-2b2 + …
�� − 1 abn-1 +
�� bn
�� =
�!�! ��� ! for n ≥ k
• So, (a+b)3 = �!�!�! a3 +
�!�!�! a2b+
�!�!�! ab2+
�!�!�!b
3 =
a3 + 3a2b+ 3ab2+ b3
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General Expansion
• The rth term in the expansion of (a+b)n is:
�� − 1 an-r+1 br-1
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Practice
• Find (a+b)6
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Solution
• Find (a+b)6
• a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6
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Example
• Find the 17th term of (a-b)23
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Solution
• Find the 17th term of (a-b)23
• ��!��! ����� ! a(23-16)b16 =
��!��! � !a
7b16
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Example
• Find the 5th term of (a-2b)3
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Solution
• Find the 3rd term of (a-2b)5
• !�! �� !a
3(2b)2 = ·�·�� a3(b)2 = 40a3b2
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Section 14.3
An Introduction to Sequences and Series
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Definitions
• A sequence is an ordered list of numbers. It can be finite,
having a fixed number of terms, or infinite, havening infinitely
many terms.
• Terms in the sequence need not be unique
• A sequence is denoted as, for example
1, 2, 3, 4, … 6, where the dots mean “and so on”
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Denoting Terms in a Sequence
• We often denote the nth term in a sequence in terms of n
an = �
���; if n = 5, a5 = 5/6
• Sometimes terms are defined recursively
an+1 = 2(3an – 4) for n > 0, a0 = 5
• We can call a sequence a function, f(n) = an
and the domain is the set of non-negative integers
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Chapter 14.4
Arithmetic Sequences and Series
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Definition
• In an arithmetic sequence the difference between sequential
terms is a fixed number, a common difference
• 8, 10, 12, 14, 16; the common difference is 2
8+2=10, 10+2=12, etc.
• We often define the nth term of a sequence in terms of n
• For the sequence given above, the nth term is
8 + (n-1)2 which is the same as 2n + 6; check it!
• This way we can find the 51st term: a51 = 102+6 = 108
• As long as we have the starting point and the common
difference we can find any term of an arithmetic sequence
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Example
Are the following Arithmetic Sequences?
If so, find their common difference
-1, -1, -1, -1, …
2, 4, 8, 16…
-1, 1, -1, 1…
3, 11/5, 7/5, 3/5…
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Solution
Are the following Arithmetic Sequences?
-1, -1, -1, -1, … yes, zero difference
2, 4, 8, 16… no; no common difference
-1, 1, -1, 1…no; no common difference
3, 11/5, 7/5, 3/5…yes, -4/5 common difference
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Example
• Find the 20th term of the sequence
7, 2, -3, -8, …
• Find the 30th term of the sequence
2/5, 4/5, 6/5, …
• Find the 15th term of the sequence
42, 1, -40, -81, …
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Solution• Find the 20th term of the sequence
7, 2, -3, -8, …
• Difference is -5, nth term is 7 + (-5)(n-1) = 12 – 5n
12- 5(20) = 12 – 100 = -88
• Find the 30th term of the sequence
2/5, 4/5, 6/5, …
• Difference is 2/5, nth term is 2/5 + (n-1)2/5 = 2/5 n
30(2/5) = 12
• Find the 15th term of the sequence
42, 1, -40, -81, …
• Difference is -41, nth term is 42 + (n-1)(-41) = 83 – 41n
83 – 41(15) = 83 – 615 = -532
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Example
• The fifth term in an arithmetic sequence is ½ and the 20th is
7/8. Find the first three terms of the sequence.
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Solution• The fifth term in an arithmetic sequence is ½ and the 20th is
7/8. Find the first three terms of the sequence.
• a + (5-1)d = ½, where a is the 1st term, and d is the common
difference
• a + (20-1)d = 7/8
• Taking the 2nd equation less the 1st, we have 15d = 3/8, or
d = 1/40
• a+4/40 = ½, from the equation for the fifth term, so a = 2/5.
Check: 2/5 + 1/10 = 5/10 = ½
2/5 + 19/40 = 16/40 + 19/40 = 35/40 = 7/8
• The first three terms are 2/5, 17/40, 9/20
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Example
• Find the common difference in an arithmetic sequence in
which the 10th term minus the 20th is 70
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Solution
• Find the common difference in an arithmetic sequence in
which the 10th term minus the 20th is 70
• Note, that since the 20th is 70 less then 10th, the difference
must be negative
• (a + 9d) – (a + 19d) = 70
• -10d = 70, d = -7
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Series
• A series is the sum of a sequence
– Finite, a fixed number of terms
– Infinite, continues to include an infinite number of terms
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Arithmetic Series
• An arithmetic series is the sum of an arithmetic sequence
• If we have a sequence 1, 3, 5, 7,
the associated arithmetic series is 1 + 3 + 5 + 7
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Famous Problem
• Find the sum of the numbers from 1 to 100
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The smart solution
• 1+99 = 100
• 2+98 = 100
• 3 + 97 = 100
.
.
.
• 49 + 51 = 100
• So we have 49 x 100, but have left out 50 and 100
• 4900 + 100 + 50 = 5050
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Finding the Sum of an Arithmetic Series
• Assume an = a + (n-1)d, where d is the common difference
and a is the first term
• The sum, Sn = (a) + (a + d) + (a + 2d) + …(a + (n-1)d)
= na + (n/2)(n-1)d = (n/2) (2a + (n-1)d) = (n/2)(2a + a + (n-1)d)
= n/2(a + an)
• The sum of an arithmetic sequence of length n is n times the
mean of the 1st and nth term
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Using for the sum of 1-100
• The sum of an arithmetic sequence of length n is n times the
mean of the 1st and nth term
• Length is 100
• Mean of 1 and 100 is 101/2 = 50.5
• 100 x 50.5 = 5050
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Examples
• Find the sum of the first 16 terms in the sequence
2, 11, 20, 29…
• Find the sum of the first 50 terms in a series whose first term
is -8 and 50th term is 139
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Solution
• Find the sum of the first 16 terms in the sequence
2, 11, 20, 29…
difference is 9, an= 2+(n-1)9; n=16 so an= 2+(15)9=137
so sum is 16[1/2(2+137)] = 1112
• Find the sum of the first 50 terms in a series whose first term
is -8 and 50th term is 139
50(-8 + 139)/2 = 25(131) = 3275
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Example
• The sum of the first 12 terms of an arithmetic sequence is
156. What is the sum of the 1st and 12th term?
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Solution
• The sum of an arithmetic sequence of length n is n times the
mean of the 1st and nth term
• The sum of the first 12 terms of an arithmetic sequence is
156. What is the sum of the 1st and 12th term?
• 156 = 12(a1 + a12)/2
• a1 + a12 = 312/12 = 26
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Example
• Find the sum of 1/e + 3/e + 5/e + …21/e
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Solution
• Find the sum of 1/e + 3/e + 5/e + …21/e
• The sum of an arithmetic sequence of length n is n times the
mean of the 1st and nth term
• The difference is 2/e; we have 11 terms
• 11(1/e + 21/e)/2 = 121/e
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A Shortcut Notation
• Sigma notation:
� ���
���Means to sum the values a, starting with n = k and ending
with n = m, ∑ ������ = ak+ak+1+…am
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Example
• ∑ 2 � �� = 24 + 25 + 26
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Example
• !�! +
!"�! +
!#�! +
!$�! …
!%"��! = ∑ !&
!�� ��
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Examples
• Find ∑ ����
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Solution
• Find ∑ ����• 1 + 2 + 3 + 4 + 5
• Sum is n( a1 + an)/2 = 5(1+5)/2 = 15
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Example
• Find ∑ (� − 1)'�������
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Solution
• Find ∑ (� − 1)'�������• 0 + 1 + 2x
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Example
• Write in sigma notation:
• 1 + 1/2 + 1/3 + 1/4 + …+1/n
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Solution
• Write in sigma notation:
• 1 + 1/2 + 1/3 + 1/4 + …+1/n
• ∑ 1/�����
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Chapter 14.5
Geometric Sequences and Series
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Definition
• A geometric sequence (or progression) is a sequence of the
form a, ar, ar2, ar3, …, where a and r are non-zero constants
(a and r can be < 0)
• r is called the common ratio
• For example, if we have a geometric series 3, 6, 12, … then
a = 3 and r = 2
• If our geometric series is 2, x, 3, …, then 3/x = x/2 so x2 = 6
a = 2 and r = ± 6 /2
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Finding the nth term
• an = arn-1
• Find the 7th term in the sequence 2, 6, 18, ...
a = 2, and r = 3, the 7th term is 2(36) = 2(729) = 1458
• As with arithmetic sequences, we can have a finite geometric
sequences
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Examples
• Find the 100th term of the sequence -1, 1, -1, 1…
• Find the 6th term of the series 1, -2 2, 8…
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Solution
• Find the 100th term of the series -1, 1, -1, 1…
The ratio is -1, so we have -1(-1)99 = 1
• Find the 6th term of the series 1, - 2, 2…
Ratio is - 2, so we have 1(- 2)n-1 = 1(- 2)5 = -4 2
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A Finite Geometric Series
• a + ar + ar2 + ar3… +arn-1 = *(��+,)
��+
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Example
• Find the sum of 1/2 + (1/2)2 + …(1/2)9
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Solution
• Find the sum of 1/2 + (1/2)2 + …(1/2)9
• Sum = %"(��
%"%-)
��%"
= %"(��
%%-"$)
��%"
= 1-1/1024 = 1023/1024
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Example
• Find ∑ ���
�����
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Solution
• Find ∑ ���
����� = 1/100 + (1/100)(1/10)+ …+ (1/100)(1/10)4
• Sum = (1/100) �� %
%-.
�� %%-
= (1/100)(10/9)[ 99,999/100,000]=
11,111/1,000,000
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Example
• Find ∑ ��
�������
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Solution
• Find ∑ ��
�������
• = (2/3) 2 + …(2/3)7 = a + ar + …ar5
• =
"#
" �� "#
/
���/� =
$0 �� /$
1"0�/� = (4/3) [
��2�����2 ]=(4/3)(665/729) =
2660/2187
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An Infinite Geometric Series
• We continue until n is infinite
• Consider ½ + 1/4 + 1/8 + …
• For n terms, the sum is
%" �� %
",
%"
= (1- ( ½ ) n)
• As n gets large, this is just 1
• The sum of an infinite series, with |r|<1 is *
��+
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Example
• Find ∑ ��
���4���
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Solution
• Find ∑ ��
���4���
• s= *��+ =
��/� = 3
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Example
• Find the sum of the infinite series 9/10 + 9/100 + 9/1000…
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Solution
• Find the sum of the infinite series 9/10 + 9/100 + 9/1000…
• Sum = a/(1-r) = (9/10) / (1-10) = 1
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Example
• Find the sum of the infinite geometric series
• -1 -�� -1/2-…
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Solution
• Find the sum of the infinite geometric series
• -1 -�� -1/2-…
• a = -1, r = ��
• Sum = -1/(1-��) = - 2/( 2-1) =(-2 2-2) /
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Repeating Decimals
• Find a fractional equivalent of 0.235353535…
• 0.2353535… = 2/10 + 35/1000 + 35/10000…
• Treat as an infinite geometric series
• Sum = 2/10 + a(1-r) = 2/10 + (35/1000)/(1-1/100)
=2/10 + 35/( 1000(99/100)) = 2/10 + 35/(990)= 233/990
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Example
• Evaluate 0.47474747…
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Solution
• Evaluate 0.47474747…
• a=47/100
• r = 1/100
• S = a/(1-r) = (47/100)/(99/100) = 47/99
83