Sequences, Series, and the Binomial Theorem

3. Infinite and Finite Sequences
An infinite sequence is a function whose domain
is the set of natural numbers {1, 2, 3, 4,…}.
2, 4, 8, 16, 32,…
terms
An finite sequence is a function whose domain
is the set of natural numbers {1, 2, 3, 4,…, n},
where n is some natural number.
2, 4, 6, 8, …, 98, 100
general term

4. Terms of a Sequences
Because a sequence is a function, it may be
described as f(n) = an = 2n , where n is a natural
number.
Example:
Write the first three terms of the sequence
whose general term is given by
an = n3 + 1.
a1 = 13 + 1 = 1 + 1 = 2 Replace n with 1.
a2 = 23 + 1 = 8 + 1 = 9 Replace n with 2.
a3 = 33 + 1 = 27 + 1 = 28 Replace n with 3.
The first three terms are 2, 9, 28, and 65.

5. Terms of a Sequences
Example:
If the general term of a sequence is given by an = 4n2
+ 5, find
a.) a10,
b.) the two-hundredth term of the sequence.
a.) a10 = 4(10)2 + 5 Replace n with 10.
= 4(100) + 5 = 405
b.) a200 = 4(200)2 + 5 Replace n with 200.
= 4(40000) + 5 = 160005

6. Terms of a Sequences
Example:
Find a general term an of the sequence whose first
five terms are
a.) 5, 10, 15, 20, 25;
b.) – 1, 2, 7, 14, 23.
a.) These numbers are the product of 5 and the first five
natural numbers, so a general term might be
an = 5n.
b.) These numbers are 2 less than the squares of the first
five natural numbers, so a general term might be
an = n2 – 2.

7. Applications Using Sequences
Example:
A mathematics lecture room has 20 rows with 8 seats in
the first row, 12 seats in the second row, 16 seats in the
third row and so on. Write an equation of a sequence
whose term corresponds to the seats in each row. How
many seats are there in the tenth row?
Since each row has 4 more seats than the previous row, the
general term would be an = 8 + 4(n – 1) where n is 1, 2, 3, …,
20.
To find the number of seats in the tenth row,
evaluate a10 = 8 + 4(10 – 1) = 8 + 4(9) = 8 + 36 = 44.

9. Arithmetic Sequences
In a sequence, when the difference of any two
consecutive terms is a constant, the sequence is an
arithmetic sequence.
5, 10, 15, 20, 25, …
d=5
The common difference is the constant, d.

10. Arithmetic Sequences
Example:
Write the first five terms of the arithmetic sequence
whose first term is 3 and whose common difference
is 4. a1 = 3
a2 = 3 + 4 = 7
a3 = 7 + 4 = 11
a4 = 11 + 4 = 15
a5 = 15 + 4 = 19
The first five terms are 3, 7, 11, 15, and 19.

11. Arithmetic Sequences
The general term, an, of an arithmetic sequence is given
by
Example: a = a + (n – 1)d
n 1
where a1 is the first term and d is the common
difference.
Consider the arithmetic sequence whose first term is 3 and common difference is
4. Write an expression for the general term an.
an = a1 + (n – 1)d
= 3 + (n – 1)4
= 3 + 4n – 4
= 4n – 1

12. Arithmetic Sequences
Example:
Find the fifth term of an arithmetic sequence whose first
three terms are 6, 11, 16.
The fifth term of the sequence is
a5 = a1 + (5 – 1)d = a1 + 4d.
a1 is the first term of the sequence, so a1 = 6.
d is the common difference of terms, so
d = a2 – a1 = 11 – 6 = 5.
Thus, a5 = a1 + 4d
= 6 + 4(5)
= 6 + 20 = 26

13. Geometric Sequences
A geometric sequence is a sequence in which each
term (after the first) is obtained by multiplying the
preceding term by a constant r.
5, 10, 20, 40, 80, …
r=2
The constant r is called the common ratio of
the sequence.

14. Geometric Sequences
Example:
Write the first five terms of a geometric
sequence whose first term is 2 and whose
common 2
a = ratio is 6.
1
a2 = 2(6) = 12
a3 = 12(6) = 72
a4 = 72(6) = 432
a5 = 432(6) = 2592
The first five terms are 2, 12, 72, 432, and 2592.

15. Geometric Sequences
In the geometric sequence whose first five terms are 2, 12, 72,
432, and 2592, notice the general pattern of the terms.
a1 = 2
a2 = 2(6) = 12 or a2 = a1(r)
a3 = 12(6) = 72 or a3 = a2(r) = (a1 ∙ r) ∙ r = a1r2
a4 = 72(6) = 432 or a4 = a3(r) = (a1 ∙ r2) ∙ r = a1r3
a5 = 432(6) = 2592 or a5 = a4(r) = (a1 ∙ r3) ∙ r = a1r4
(subscript – 1) is the power

16. Geometric Sequences
The general term, an, of a geometric sequence is given by
an = a1 rn – 1
where a1 is the first term and r is the common ratio.
Example:
Find the fifth term of the geometric sequence
1
whose first term is 6 and whose common ratio is
.
n 1 2
1
an = a1rn – 1 6 2
5 1 4
1 1 1 6 3
a5 6 6 6
2 2 16 16 8

17. Geometric Sequences
Example:
Find the ninth term of the geometric sequence
whose first three terms are 3, –12, 48.
Since r is the common ratio of the terms,
a2 12
r 4.
a1 3
a9 = a1r9 – 1
= 3(– 4)8
= 196,608

18. Geometric means are the terms between any two
nonconsecutive terms of a geometric sequence.

19. Example 4: Finding Geometric Means
Find the geometric mean of and .
Use the formula.

20. Check It Out! Example 4
Find the geometric mean of 16 and 25.
Use the formula.

23. A Harmonic Progression is a
sequence of quantities whose
reciprocals form an
arithmetic progression.

24. * The series formed by the
reciprocals of the terms of a
geometric series are also
geometric series.

25. * There is no general method
of finding the sum of a
harmonic progression.

26. Example
The Sequence
“s1 , s2 , … , sn”
is a Harmonic Progression if
“1/s1 , 1/s2 , … , 1/sn”
forms an Arithmetic Progression.

27. Method For
Re-checking a
Harmonic
Progression

28. A Harmonic Progression is a set of
values that, once reciprocated, results
to an Arithmetic Progression. To check ,
the reciprocated values must possess a
rational common difference. Once this
has been identified, we may say that
the sequence is a Harmonic
Progression.

29. Harmonic Means are the
terms found in between
two terms of a harmonic
progression.

30. Determine which of the
following are Harmonic
Progressions.

31. 1) 1 ,1/2 , 1/3 , 1/4 , ...

32. Step 1: Reciprocate all the given terms.
* The reciprocals are: 1 , 2 , 3 , 4 , …
Step 2: Identify whether the reciprocated
sequence is an Arithmetic Progression by checking
if a common difference exists in the terms.

33. Answer: It is a Harmonic
Progression.

34. 2) 1 , 1/4 , 1/5 , 1/7 , ...

35. Step 1: Reciprocate all the given terms.
* The reciprocals are: 1 , 4 , 5 , 7 , …
Step 2: Identify whether the reciprocated
sequence is an Arithmetic Progression by
checking if a common difference exists in
the terms.

36. Answer: It is NOT a
Harmonic Progression.

37. Determine the next three
terms of each of the
following Harmonic
Progressions.

38. 1) 24 , 12 , 8 , 6 , …

39. Solution:
24 , 12 , 8 , 6 , …
= 1/24 , 1/12 , 1/8 , 1/6
* To find the common difference:
1/12 – 1/24
= 2/24 – 1/24
= 1/24

40. You can subtract the
second term to the first
term, the third to the
second term, the forth to
the third term, and so on
and so forth.

41. To get the next three
terms:
5th Term = 1/6 + 1/24
= 4/24 + 1/24
= 5/24
* Reciprocate
= 24/5

42. 6th Term = 5/24 + 1/24
= 6/24
= 1/4
* Reciprocate
=4

43. 7th Term = 1/4 + 1/24
= 6/24 + 1/24
= 7/24
* Reciprocate
= 24/7

44. Find the Harmonic Mean
between the following
terms.

45. 1) 12 and 8

46. Step 1: Reciprocate all the
given terms.
* The reciprocals are: 1/12
and 1/8
Step 2: Arrange the given
terms as follows:

47. 1/12 Harmonic Mean 1/8
1’st term 2’nd term 3’rd term

48. *For this problem, we will
use the formula:
tn = t1 + (n – 1)d

49. We may now substitute
the values in the problem
to the formula to find the
common difference (d)
and the Harmonic Mean
as follows:

50. t3 = t1 + (3 - 1)d
1/8 = 1/12 + 2d
1/8 – 1/12 = 2d
(3 – 2) / 24 = 2d
(3 – 2) = 48d
1 = 48d
d = 1/48

51. *After getting the
Common Difference, add
it to the first term to get
the Harmonic Mean
between the two terms.

52. t2 = t1 + d
= 1/12 + 1/48
= (4 + 1) / 48
= 5/48
*Reciprocate
= 48/5

53. Insert three Harmonic
Means between the
following terms:

54. 1) 36 and 36/5

55. Step 1: Reciprocate all the
given terms.
* The reciprocals are: 1/36
and 5/36
Step 2: Arrange the given
terms as follows:

56. 1’st term
Harmonic Means
2’nd , 3’rd , and 4’th term
5/36
5’th term

57. *For this problem, we will
use the formula:
tn = t1 + (n – 1)d

58. We may now substitute
the values in the problem
to the formula to find the
common difference (d)
and the Harmonic Means
as follows:

59. t5 = t1 + (5 - 1)d
5/36 = 1/36 + 4d
5/36 – 1/36 = 4d
(5 - 1) / 36 = 4d
(5 - 1) = 144d
4 = 144d
d = 4/144
= 1/36

60. *After getting the
Common Difference, add
it to the first term, then
add it to the second term,
and then add it to the
third term to get the
Harmonic Means between
the two terms.

61. t2 = t1 + d
= 1/36 + 1/36
= 2/36
= 1/18
*Reciprocate
= 18

62. t3 = t2 + d
= 2/36 + 1/36
= 3/36
= 1/12
*Reciprocate
= 12

63. t4 = t3 + d
= 3/36 + 1/36
= 4/36
= 1/9
*Reciprocate
=9

64. Therefore, the three
means between 36 and
36/5
are 18, 12, and 9.

65. Activity

66. Determine if the following
are harmonic
progressions or not:
1) 1/12 , 1/24 , 1/36
2) 2 , 5 , 7 , 8
3)1/5 , 1/10 , 1/15

67. Find the next three terms
in the following harmonic
progressions:
1) 1/2 , 1/5 , 1/8 , 1/11 , …
2) 19 , 17 , 15 , 13, …
3) 12 , 6 , 4 , 3 , …

68. Find the harmonic mean
between:
1) 1/2 and 1/5
2) 1 and 1/9

69. Insert three harmonic
means between:
1) 1/2 and 1/8
2) 1 and 1/10

71. Finite Series
A sum of the terms of a sequence is called a series.
A series is a finite series if it is the sum of a
finite number of terms.
Sequence Series
2, 4, 6, 8 2 + 4 + 6 + 8 = 20
5, 10, 20, 40 5 + 10 + 20 + 40 = 75

72. Infinite Series
A series is an infinite series if it is the sum of
all the terms of the sequence.
Sequence Series
2, 4, 6, 8, … 2+4+6+8+…
5, 10, 20, 40, … 5 + 10 + 20 + 40 + …
When the general term of a sequence is known,
summation notation is used for denoting a series.
The Greek uppercase letter sigma, Σ, is used
to mean “sum.”

73. Infinite Series
3
The expression (2n 1) is read “the sum of 2n –
n 1
1 as n goes from 1 to 3.” This expression means
the sum of the first three terms of the sequence
whose general term is an = 2n – 1.
Often the variable i is used instead of n.
3
(2i 1) (2 1 1) (2 2 1) (2 3 1)
i 1
index of 1 3 5
summation 9

74. Infinite Series
Example:
5
Evaluate (n2 2).
i 0
5
(n 2
2) 02 2 12 2 22 2 32 2
i 0
2 2
4 2 5 2
2 3 6 11 18 27
67

75. Partial Sums
The sum of the first n terms of a sequence is
a finite series known as a partial sum, Sn.
S1 = a1
S2 = a1 + a2
S3 = a1 + a2 + a3
In general, Sn is the sum of the first n terms of a
sequence.
n
Sn an
i 1

76. Partial Sums
Example:
Write the series using summation notation.
4 + 10 + 16 + 22 + 28
Since the difference of each term and the preceding
term is 6, this is an arithmetic sequence with a1 = 4
and d = 6. a + (n – 1)d = 4 + (n – 1)6
a =
n 1
= 4 + (n – 1)6
= 4 + 6n – 6
= 6n – 2
5
4 + 10 + 16 + 22 + 28 6i 2
i 1

77. Partial Sums
Example:
Write the series using summation notation.
3 + 9 + 27 + 81 + 243
Since each term is the product of the preceding term
and 3, this is a geometric sequence with a1 = 3 and r = 3.
an = a1rn – 1 = 3(3)n – 1
= 3 13 n – 1
= 31 + (n – 1)
= 3n
5 i
3 + 9 + 27 + 81 + 243 3
i 1

78. Example:
Find the sum of the first four terms of the sequence
5 n
whose general term is n
a .
3n
45 i 5 1 5 2 5 3 5 4
S4
i 1 3i 31 3 2 3 3 3 4
4 3 2 1
3 6 9 12
48 18 8 3
36 36 36 36
77 5
2
36 36

80. Partial Sums of Arithmetic
Sequences
The partial sum Sn of the first n terms of an arithmetic sequence
is given by
n
Sn (a1 an)
2
where a1 is the first term of the sequence and an is the nth term.

81. Example:
Find the sum of the first four terms of the
arithmetic sequence 3, 9, 15, 21, 27, …
n
Sn (a1 an)
2
4
S4 (3 21)
2
2(24)
48

82. Example:
Find the sum of the first 25 even integers.
Because 2, 4, 6, …, 50 is an arithmetic sequence, the
formulas for Sn is used with n = 25, a1 = 2, and an = 50.
n
Sn (a1 an)
2
25
S25 (2 50)
2
25
(52)
2
650

83. The partial sum Sn of the first n terms of a
geometric sequence is given by
a1(1 r n)
Sn
1 r
where a1 is the first term of the sequence, r is
the common ratio, and r 1.

84. Example:
Find the sum of the first five terms of the
geometric sequence 3, 12, 48, 192, 768, 3072, …
a1(1 r n)
Sn
1 r
12
3(1 45) r 4
S5 3
1 4
3(1 1024)
3
3( 1023)
3
1023

85. Example:
Chelsea made P20,000 during the first year she was self-
employed. She made an additional 15% more than the
previous year in each subsequent year.
a.) How much did she make during her fifth year of
business?
b.) What were her total earnings during the five
years?
Chelsea’s earnings are modeled by a geometric
sequence where n = 5, a1 = 20,000, and r = 1.15
a.) an = a1rn – 1
a5 = 20,000(1.15)4 34,980.13 Continued.

86. Example continued:
b.) S a1(1 r n)
n 1 r
20,000(1 1.155)
S5
1 1.15
20,000(1 2.0114)
0.15
20,000( 1.0114) 134,853.33
0.15
Chelsea made approximately P34,980.13 during her
fifth year of self-employment, and a total of
P134,853.33 during the first five years.

87. Infinite Geometric Sequences
The sum S∞ of the terms of an infinite geometric
sequence is given by
a1
S
1 r
where a1 is the first term of the sequence, r
is the common ratio, and |r| < 1. If |r|
1, S does not exist.

88. Example:
Find the sum of the terms of the geometric
sequence 4, 8 , 16 , 32 ,
3 9 27
a1
S
1 r
8 2
r 4 4
3 3
2
1
3
4
12
1
3

90. Expanding Binomials
Expanding a binomial such as (a + b)n means
to write the factored form as a sum.
(a + b)0 = 1 1 term
(a + b)1 = a + b 2 terms
(a + b)2 = a2 + 2ab + b2 3 terms
(a + b)3 = a3 + 3a2b + 3ab2 + b3 4 terms
(a + b)4 = a4 + 4a3b + 6a2b2 + 4a1b3 + b4 5 terms
(a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5 6 terms

91. Expanding Binomials
(a + b)0 = 1
(a + b)1 = a + b
(a + b)2 = a2 + 2ab + b2
(a + b)3 = a3 + 3a2b + 3ab2 + b3
(a + b)4 = a4 + 4a3b + 6a2b2 + 4a1b3 + b4
(a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5
1. The expansion of (a + b)n contains n + 1 terms.
2. The first term is an and the last term is bn.
3. The powers of a decrease by 1 for each term; the powers of b
increase by 1 for each term.
4. The sum of the exponents of a and b is n.

92. Pascal’s Triangle
There are also patterns in the coefficients of the terms.
When written in a triangular array, the coefficients are
called Pascal’s triangle.

93. Pascal’s Triangle
(a + b)0 1 n=0
(a + b)1 1 1 n=1
(a + b)2 1 2 1 n=2
(a + b)3 1 3 3 1 n=3
(a + b)4 1 4 6 4 1 n=4
(a + b)5 1
1 5 10 10 5 1 n=5
1 6
6 15 20 15 6 1
Add the consecutive numbers in the row for n =
5 and write each sum “between and below” the
pair.

94. Pascal’s Triangle
Example:
Expand (a + b)7.
Use n = 7 row of Pascal’s triangle as the coefficients
and the noted patterns.
1 6 15 20 15 6 1 n=6
1 7 21 35 35 21 7 1 n=7
(a + b)7 = 1a7 + 7a6b + 21a5b2 + 35a4b3 + 35a3b4
+ 21a2b5 + 7ab6 + 1b7

95. Factorials
An alternative method for determining the
coefficients of (a + b)n is based on using factorials.
The factorial of n, written n! (read “n factorial”), is
the product of the first n consecutive natural
numbers.
Factorial of n: n!
If n is a natural number, then
n! = n(n – 1)(n – 2)(n – 3) . . . ∙ 3 ∙ 2 ∙ 1.
The factorial of 0, written 0!, is defined to be 1.

96. Evaluating Factorials
Example:
Evaluate each expression.
a.) 6!
8!
b.)
3!
a.) 6! 6 5 4 3 2 1 720
8! 8 7 6 5 4 3 2 1
b.) 6720
3! 3 21

97. Binomial Theorem
It can be proved that the coefficients of terms in the expansion of (a
+ b)n can be expressed in terms of factorials. Following the earlier
patterns and using the factorial expressions of the coefficients, we
have the binomial theorem.
Binomial Theorem
If n is a positive integer, then
n n 1 1 n(n 1) n 2 2
(a b) n an a b a b
1! 2!
n(n 1)(n 2) n 3 3
a b  bn
3!

98. Binomial Theorem
Example:
Use the binomial theorem to expand (x + 3)4.
4 4 4 3 4 3 2 2 4 3 2 3 4
(x 3) x x3 x3 x3 3
1! 2! 3!
x4 4 3x3 6 9x2 4 27x 81
x4 12x3 54x2 108x 81

99. Binomial Theorem
Example:
Use the binomial theorem to expand (3a – 5b)6.
6 6 6 5 6 5 4 2
(3a 5b) (3a) (3a) ( 5b) (3a) ( 5b)
1! 2!
6 5 4 6 5 4 3
(3a)3( 5b)3 (3a) 2( 5b) 4
3! 4!
6 5 4 3 2
(3a)( 5b)5 ( 5b)6
5!
729a 6 7290a5b 30,375a 4b 2 67,500a 3b3
84,375a 2b 4 56, 250ab5 15,625b6

100. Binomial Expansion
(r + 1)st Term in a Binomial Expansion
The (r + 1)st term of the binomial expansion of
n!
(a + b)n is a n rb r.
r !(n r)!

101. Binomial Expansion
Example:
Find the ninth term in the expansion of (3x – 5y)10.
n = 10, a = 3x, b = – 5y, r + 1 = 9, therefore r = 8
n! 10!
a n rbr (3x)10 8( 5 y)8
r !(n r)! 8!(10 8)!
45 9x 2 390,625 y 8 158, 203,125x 2 y 8