The document discusses sequences and summations. It provides examples and definitions of different types of sequences such as arithmetic and geometric sequences. It also discusses recurrence relations, which express a term in a sequence based on prior terms. Examples are provided to demonstrate finding terms of sequences given a recurrence relation. The document is a lecture on discrete mathematics concepts related to sequences.

1. Discrete Structures
Sequences and Summations
Dr. Muhammad Humayoun
Assistant Professor
COMSATS Institute of Computer Science, Lahore.
mhumayoun@ciitlahore.edu.pk
https://sites.google.com/a/ciitlahore.edu.pk/dstruct/
A lot of material is taken from the slides of Dr. Atif and Dr. Mudassir
1

2. 2

3. Sequence
• Sequence: a discrete structure used to represent
an ordered list of elements e.g.:
– 1, 2, 3, 5, 8 is a sequence with five terms
– 1, 3, 9, 27, 81 , . . . , 3n, . . . (infinite sequence)
Definition 1: A sequence is a function from a
subset of Z to some set S.
• The notation an denotes the image of the integer n
• an : a term of the sequence
• {an} : entire sequence
– Same notation as sets!
3
Z S
n an

4. Length of Sequence:
A sequence may be Finite or Infinite.
4

5. Sequences and Rules
5

6. Naming Sequences
6

7. Examples 1
at Page# 156
• Consider the sequence {an}, where an = 1/n.
The list of the terms of this sequence beginning with a1:
a1, a2, a3, a4, …
{1, 1/2, 1/3, 1/4 , … }
• Consider the sequence {an}, where an = 3n.
The list of the terms of this sequence beginning with a1:
{3, 6, 9, 12 , …}
• The Sequence {bn}, where bn=2n
The list of the terms of this sequence beginning with b1:
{2, 4, 8, 16, … }
7

8. Example 3
at Page# 157
• Is the sequence {sn} with sn = −1 + 4n an arithmetic
progressions with initial terms -1 and common
differences equal to 4? if we start at n = 0. The list of
terms s0 , s1 , s2 , s3 , . . . begins with −1, 3, 7, 11, . . . ,
Yes, it is arithmetic progression.
• Is the sequence {tn} with tn = 7 − 3n an arithmetic
progressions with initial terms 7 and common
differences equal to -3? if we start at n = 0. the list of
terms t0 , t1 , t2 , t3 , . . . begins with 7, 4, 1,−2, . . . .
 Yes, it is arithmetic progression.
8

9. Types of Sequence
• Arithmetic Sequence
Next term is calculated by adding in the preceding term.
• Geometric Sequence
 Next term is calculated by multiplying by the preceding term.
Arithmetic Series
 Sum of Arithmetic Sequences
Geometric Series
 Sum of Geometric Sequences
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10. • Next class
10

11. Types of Sequence and Series
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12. Arithmetic Progression/ Sequences
12

13. Arithmetic Progression/ Sequences
(Conti…)
13

14. Sequences and Summations
14

15. Sequences and Summations
Arithmetic Progressions
15

16. Sequences and Summations
Arithmetic Progressions
16

17. Sequences and Summations
Arithmetic Progressions
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18. Arithmetic Progressions
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19. Arithmetic Progressions
19

20. Arithmetic Progressions
20

21. Arithmetic Progressions
21

22. General form of an Arithmetic Sequence
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23. 23

24. 24

25. 25

26. 26

27. 27

28. 28

29. 29

30. 30

31. Geometric Progression
• A geometric progression is a sequence of the form
a, ar1, ar2, . . . , arn-1, . . .
where
the initial term a
and the common ratio r are real numbers.
• A geometric progression is a discrete analogue of
the exponential function f (x) = arx.
• Geometric Progression: an = ar n-1
 ar0, ar1, ar2, ar3, …
• Is the sequence {bn} Geometric Progression???
31

32. Example# 2
at Page 157
• Is the sequence {bn} with bn = (−1)n a geometric progression with
initial term 1 and common ratio −1? if we start at n = 0. the list of
terms b0 , b1 , b2 , b3 , b4 , . . . begins with 1,−1, 1,−1, 1, . . .
 Yes, it is Geometric Progression.
• Is the sequence {cn} with cn = 2 ・ 5n a geometric progression
with initial term 2 and common ratio 5? if we start at n = 0. the
list of the terms c0 , c1 , c2 , c3 , c4 , . . . begins with 2, 10, 50, 250,
1250, . . .
 Yes, it is Geometric Progression.
• Is the sequence {dn} with dn = 6 ・ (1/3)n a geometric progression
with initial term 6 and common ratio 1/3? if we start at n = 0. the list
of the terms d0 , d1 , d2 , d3 , d4 , . . . begins with 6, 2, 2/3, 2/9,
2/27,...
 Yes, it is Geometric Progression.
32

33. Strings
• Finite Sequences of the form a1, a2, . . . , an are
called strings.
• The length of a string is the number of terms in
the string.
• EXAMPLE 4 at Page# 157, The string abcd is a
string of length four.
• The empty string, denoted by λ, is the string that
has no terms. The empty string has length zero.
33

34. Recurrence Relations
A recurrence relation for the sequence {an} is an
equation that expresses an in terms of one or more
of the previous terms of the sequence.
A sequence is called a solution of a recurrence
relation if its terms satisfy the recurrence relation.
(A recurrence relation is said to recursively define a
sequence.)
34

35. Example 5 at Page# 158
Let {an} be a sequence that satisfies the recurrence
relation an = an−1 + 3 for n = 1, 2, 3, ...
Suppose that a0 = 2. What are a1, a2, and a3?
• a1 = a0 + 3 = 5
• a2 = a1 + 3 = 8
• a3 = a2 + 3 = 11
{5, 8, 11, …}
35

36. Example 6 at Page# 158
Let {an} be a sequence that satisfies the recurrence
relation an = an-1 − an-2 for n = 2, 3, 4, . . . , Suppose
that a0 = 3 and a1 = 5 . What are a2, and a3?
• an = an-1 − an-2
• a2 = a2-1 − a2-2 a2 = a1 − a0
• a2 = a1 − a0 a2 = 5 − 3= 2
• a3 = a2 − a1 a3 = 2 − 5= -3
{2,-3, …}
36

37. Fibonacci sequence
EXAMPLE 7 at Page# 158
• The Fibonacci sequence, f0 , f1 , f2 , . . . , is defined by
the initial conditions f0 = 0, f1 = 1, and the recurrence
relation fn = fn−1 + fn−2 for n = 2, 3, 4, . . .
• Find the Fibonacci numbers f2 , f3 , f4 , f5 ,and f6?
• fn = fn−1 + fn−2
• f2 = f2−1 + f2−2 f2 = f1 + f0 f2 = 1 + 0 f2 = 1
• f3 = f3−1 + f3−2 f3 = f2 + f1 f3 = 1 + 1 f3 = 2
• f4 = f4−1 + f4−2 f4 = f3 + f2 f4 = 2 + 1 f4 = 3
• f5 = f5−1 + f5−2 f5 = f4 + f3 f5 = 3 + 2 f5 = 5
• f6 = f6−1 + f6−2 f6 = f5 + f4 f6 = 5 + 3 f6 = 8
37

38. EXAMPLE 8 at Page# 159
• Suppose that {an} is the sequence of integers defined by an = n!,
the value of the factorial function at the integer n, where n = 1, 2,
3, . . .. Because n! = n((n − 1)(n − 2) . . . 2 ・ 1) = n(n − 1)! = nan−1,
we see that the sequence of factorials satisfies the recurrence
relation an = nan−1, together with the initial condition a1 =1.
38

39. Example 9
at Page#159
• Determine whether the sequence {an}, where an
= 3n for every nonnegative integer n, is a solution
of the recurrence relation an = 2an−1 − an−2 for n =
2, 3, 4, . . . .
• Solution
For n  2 we see that
2an-1 – an-2 = 2(3(n – 1)) – 3(n – 2) = 3n = an.
Therefore, {an} with an=3n is a solution of the
recurrence relation.
39

40. Example 9
at Page#159
• Determine whether the sequence {an}, where an
= 2n for every nonnegative integer n, is a solution
of the recurrence relation an = 2an−1 − an−2 for n =
2, 3, 4, . . . . Answer the same question where an
= 2n
•For n  2 we see that
2an-1 – an-2 = 2(2(n – 1)) – 2(n – 2) = 2n = an.
•Therefore, {an} with an=2n is a solution of the
recurrence relation.
40

41. Example 9
at Page#159
• Determine whether the sequence {an}, where an =
3n for every nonnegative integer n, is a solution of
the recurrence relation an = 2an−1 − an−2 for n = 2, 3,
4, . . . . Answer the same question where an = 2n
• For n  2 we see that
• Note that a0 = 1, a1 = 2, and a2 = 4.
• Because 2a1 − a0 = 2 ・ 2 − 1 = 3 ≠ a2
•Therefore, {an} with an= 2n is not a solution of the
recurrence relation.
41

42. Example 9
at Page#159
• Determine whether the sequence {an}, where an
= 3n for every nonnegative integer n, is a solution
of the recurrence relation an = 2an−1 − an−2 for n =
2, 3, 4, . . . . Answer the same question where an
= 5
•For n  2 we see that
2an-1 – an-2 = 2(5) – 5 = 5 = an.
•Therefore, {an} with an=5 is a solution of the
recurrence relation.
42

43. Example 10 at Page# 159
Not Included
Solve the recurrence relation and initial condition in Example 5.
Let {an} be a sequence that satisfies the recurrence relation an = an−1 + 3 for n =
1, 2, 3, ... Suppose that a0 = 2. What are a1, a2, and a3?
• a1 = a0 + 3 = 5 a1 = 2 + 3 = 5 a1 = 2 + 3 = 2 + 3 .1
• a2 = a1 + 3 = 8 a2 = 5 + 3 = 8 a2 = (2 + 3 ) + 3 = 2+3.2
• a3 = a2 + 3 = 11 a3 = 8 + 3 = 11 a3 = [(2 + 3 ) + 3] + 3 = 2+3.3
• ...
• an = an−1 + 3 = 2 + 3(n − 1).
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44. 44

45. Determining the Sequence formula
• 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, …
– The sequence alternates 1’s and 0’s, increasing the
number of 1 and 0 each time.
• 1, 2, 2, 3, 4, 4, 5, 6, 6, 7, 8, 8, …
– The sequence increases by one but repeats all even
numbers twice.
• 1, 0, 2, 0, 4, 0, 8, 0, 16, 0, …
– The non-zero numbers are geometric sequences (2n),
such that every non-zero number is succeeded by a
zero.
45

46. Determining the Sequence formula
• 3, 6, 12, 24, 48, 96, 192, …
Each term is twice the previous
a1 = 3, an = 2*an-1 (recurrence relation)
• Is the above a Geometric Sequence? (an = arn-1)?
an = 3 * 2n-1 (sequence formula)
• 15, 8, 1, -6, -13, -20, -27, ….
Each term is 7 less than previous term
a1 = 15, an = an-1 - 7
an = 22 – 7*n
46

47. Example 12
at Page# 161
• Find formulae for the sequences with the following first five terms:
• (a) 1, 1/2, 1/4, 1/8, 1/16
• Solution:
 The sequence with an = (½)n, n = 0, 1, 2, . . . is a possible match. It is a
geometric progression with
 a = 1 and r = 1/2.
• (b) 1, 3, 5, 7, 9
• Solution:
 We note that each term is obtained by adding 2 to the previous term. The
sequence with an = 2n + 1, n = 0, 1, 2, . . . is a possible match. It is a
Arithmetic progression with
 a = 1 and d = 2.
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48. Example 12
at Page# 161
• (c) 1, −1, 1, −1, 1.
• Solution:
The terms alternate between 1 and −1. The sequence
with an = (−1)n, n = 0, 1, 2 . . . is a possible match. It is
a geometric progression with
a =1 and r = −1.
48

49. Example 13
at Page# 161
• How can we produce the terms of a sequence if
the first 10 terms are 1, 2, 2, 3, 3, 3, 4, 4, 4, 4?
• Solution:
A rule for generating this sequence is that the integer
n appears exactly n times.
49

50. Example 14
at Page# 161
• How can we produce the terms of a sequence if
the first 10 terms are 5, 11, 17, 23, 29, 35, 41, 47,
53, 59?
• Solution:
A rule for generating this sequence is 5+6n
It is an arithmetic progression with a = 5 and d = 6.
50

51. Example 15
at Page# 161
• How can we produce the terms of a sequence if the
first 10 terms are 1, 3, 4, 7, 11, 18, 29, 47, 76, 123?
• Solution:
• We guess that the sequence is determined by the
recurrence relation
Ln = Ln−1 + Ln−2 with initial conditions L1 = 1 and L2 = 3
• Note:
It is a Same recurrence relation as the Fibonacci
sequence, but with different initial conditions). This
sequence is known as the Lucas sequence.
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52. Example 16
at Page# 161
• Conjecture a simple formula for an if the first 10
terms of the sequence {an} are 1, 7, 25, 79, 241,
727, 2185, 6559, 19681, 59047.
• Solution:
We see that an = 3n − 2 for 1 ≤ n ≤ 10 and conjecture
that this formula holds for all n.
52

53. Some Useful Sequences
n(n+1)/2 : 1, 3, 6, 10, 15, …
53

54. Determining the Sequence formula
• 3, 5, 8, 12, 17, 23, 30, 38, 47, ….
– Initial condition: a1 = 3
– a2 = 5, a3 = 8, a4 = 12
– a2 = a1 + 2
– a3 = a2 + 3
– a4 = a3 + 4
– ….
– an = an-1 + n
– The difference between successive terms increased
by 1
54

55. Determining the Sequence formula
55
See Examples 1 to 16 on pages 156 - 162

56. Summations
• The sum of the terms am , am+1 , . . . , an from the
sequence {an}is:
am+ am+1+ . . . +an
denoted by or
Where,
 denotes Summation
j is the index of summation
m is lower limit
n is upper limit
56


n
m
j
j
a

57. Sequences and Summations
57

58. 58

59. EXAMPLE 17
at Page# 163
• Use summation notation to express the sum of
the first 100 terms of the sequence {aj }, where
aj = 1/j for j = 1, 2, 3, . . . .
• Solution:
59

60. Example 18
at Page# 164
• What is the value of ?
60

61. Example 19
at Page# 164
61
What is the value of ?

62. Example 20
at Page# 164
What is value of by shifting Index of summation ?
Note: We want the index of summation to run between 0 to 4
rather than from 1 to 5.
• Let , k= j-1 (for shifting indexes of summations)
• then by putting the values of limits
for j=1, k=j-1 k=1-1=0
for j=5, k=j-1 k=5-1=4
• By putting the value of j2, we have if k=j-,1 then j=k+1
becomes
By shifting k to j, we have
• = (0+1)2+ (1+1)2+ (2+1)2+ (3+1)2+ (4+1)2 = 1+4+9+16+25=55
62

63. 63

64. 64

65. See Ex. 17-20,
22, starting
from Page
163.

66. Examples
1. Express the sum of first 100 terms of the sequence
{𝑎𝑛} where 𝑎𝑛 =
1
𝑛
for n=1, 2, 3, ….
2. What is the value of 𝑗=1
5
𝑗2
3. What is the value of 𝑘=4
8
−1 𝑘
66

67. 67

68. Example 21
at Page# 165
• Evaluate the following double Summation
68
=6(1)+ 6(2)+ 6(3)+6(4)
=6+12+18+24
= 60

69. Example 22
at Page# 165
• What is the value of ?
• Solution:
• represents the sum of the values of s
for all the members of the set {0, 2, 4}, it follows
that
= 0+2+4= 6
69

70. Example 23 at Page# 166
70
Find

71. 71

72. Exercise
at page#167
72

73. Exercise
at page#167
73

74. Exercise
at page#167
74

75. Exercise
at page#167
75

76. Exercise
at page#167
76
2, 5, 8, 11, 14, 17, 20, 23, 26, 29.

77. Exercise
at page#167
77

78. Exercise
at page#167
78
2,4,6,10,16,26,42,68,110,178

79. Exercise
at page#167
79

80. Exercise
at page#167
80

81. Exercise
at page#167
81

82. Exercise
at page#167
82

83. Exercise
at page#167
83

84. Exercise
at page#167
84

85. Exercise
at page#167
85

86. Exercise
at page#167
86

87. Exercise
at page#167
• 1, 2, 4, 7, 11, 16, 22, ....
• 1, 2, 4, 1, 2, 4, 1, 2, 4, ....
• 1, 2, 4, 8, 16, 32
87