chapter 14 14...however • consider (a+b)5 • the first coefficient is 1 • the second...

11

Chapter 14

Additional Topics in Algebra

Chapter 14.1

Mathematical Induction

2

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

3

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

4

Example

• Show 2 + 4 + 6 +…+ 2n = n(n+1)

5

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

6

Example

• Show 23 + 43 + 63 + (2n) 3 = 2n2(n+1) 2

7

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

8

Example

• Prove 5 + 9 + 13 +…(4n+1) = n(2n+3)

9

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

10

Example

• Prove 1x2 + 3x4 + 5x6 + (2n-1)(2n) = n(n+1)(4n-1)/3

11

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

12

Chapter 14.2

The Binomial Theorem

13

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

14

Pascal’s Triangle

• You can always find the coefficients by writing this triangle

15

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

16

Factorial

• n! = 1 x 2 x 3 x 4 x …n

• 1! = 1

• 0! = 1 (by definition)

• Find (n+1)!/(n-1)!

17

Example

• Find (n+1)!/(n-1)! = n(n+1)= n2 + n

18

The Binomial Coefficient

• �� =

�!�! ��� ! for n ≥ k

• 53 =

!�!�!

·�·�·�·�(�·�·�)(�·�) =

·��·� = 10

19

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

20

General Expansion

• The rth term in the expansion of (a+b)n is:

�� − 1 an-r+1 br-1

21

Practice

• Find (a+b)6

22

Solution

• Find (a+b)6

• a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6

23

Example

• Find the 17th term of (a-b)23

24

Solution

• Find the 17th term of (a-b)23

• ��!��! ����� ! a(23-16)b16 =

��!��! � !a

7b16

25

Example

• Find the 5th term of (a-2b)3

26

Solution

• Find the 3rd term of (a-2b)5

• !�! �� !a

3(2b)2 = ·�·�� a3(b)2 = 40a3b2

27

Section 14.3

An Introduction to Sequences and Series

28

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”

29

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

30

Chapter 14.4

Arithmetic Sequences and Series

31

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

32

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…

33

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

34

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, …

35

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

36

Example

• The fifth term in an arithmetic sequence is ½ and the 20th is

7/8. Find the first three terms of the sequence.

37

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

38

Example

• Find the common difference in an arithmetic sequence in

which the 10th term minus the 20th is 70

39

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

40

Series

• A series is the sum of a sequence

– Finite, a fixed number of terms

– Infinite, continues to include an infinite number of terms

41

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

42

Famous Problem

• Find the sum of the numbers from 1 to 100

43

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

44

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

45

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

46

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

47

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

48

Example

• The sum of the first 12 terms of an arithmetic sequence is

156. What is the sum of the 1st and 12th term?

49

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

50

Example

• Find the sum of 1/e + 3/e + 5/e + …21/e

51

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

52

A Shortcut Notation

• Sigma notation:

� ���

���Means to sum the values a, starting with n = k and ending

with n = m, ∑ ������ = ak+ak+1+…am

53

Example

• ∑ 2 � �� = 24 + 25 + 26

54

Example

• !�! +

!"�! +

!#�! +

!$�! …

!%"��! = ∑ !&

!�� ��

55

Examples

• Find ∑ ����

56

Solution

• Find ∑ ����• 1 + 2 + 3 + 4 + 5

• Sum is n( a1 + an)/2 = 5(1+5)/2 = 15

57

Example

• Find ∑ (� − 1)'�������

58

Solution

• Find ∑ (� − 1)'�������• 0 + 1 + 2x

59

Example

• Write in sigma notation:

• 1 + 1/2 + 1/3 + 1/4 + …+1/n

60

Solution

• Write in sigma notation:

• 1 + 1/2 + 1/3 + 1/4 + …+1/n

• ∑ 1/�����

61

Chapter 14.5

Geometric Sequences and Series

62

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

63

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

64

Examples

• Find the 100th term of the sequence -1, 1, -1, 1…

• Find the 6th term of the series 1, -2 2, 8…

65

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

66

A Finite Geometric Series

• a + ar + ar2 + ar3… +arn-1 = *(��+,)

��+

67

Example

• Find the sum of 1/2 + (1/2)2 + …(1/2)9

68

Solution

• Find the sum of 1/2 + (1/2)2 + …(1/2)9

• Sum = %"(��

%"%-)

��%"

= %"(��

%%-"$)

��%"

= 1-1/1024 = 1023/1024

69

Example

• Find ∑ ���

�����

70

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

71

Example

• Find ∑ ��

�������

72

Solution

• Find ∑ ��

�������

• = (2/3) 2 + …(2/3)7 = a + ar + …ar5

• =

"#

" �� "#

/

���/� =

$0 �� /$

1"0�/� = (4/3) [

��2�����2 ]=(4/3)(665/729) =

2660/2187

73

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 *

��+

74

Example

• Find ∑ ��

���4���

75

Solution

• Find ∑ ��

���4���

• s= *��+ =

��/� = 3

76

Example

• Find the sum of the infinite series 9/10 + 9/100 + 9/1000…

77

Solution

• Find the sum of the infinite series 9/10 + 9/100 + 9/1000…

• Sum = a/(1-r) = (9/10) / (1-10) = 1

78

Example

• Find the sum of the infinite geometric series

• -1 -�� -1/2-…

79

Solution

• Find the sum of the infinite geometric series

• -1 -�� -1/2-…

• a = -1, r = ��

• Sum = -1/(1-��) = - 2/( 2-1) =(-2 2-2) /

80

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

81

Example

• Evaluate 0.47474747…

82

Solution

• Evaluate 0.47474747…

• a=47/100

• r = 1/100

• S = a/(1-r) = (47/100)/(99/100) = 47/99

83