This document discusses sequences and their properties. It defines sequences and provides examples of different types of sequences including arithmetic, geometric, harmonic, and Fibonacci sequences. It discusses finding formulas for the nth term of sequences and calculating sums of sequence terms. Examples are provided to demonstrate finding sequence terms, recognizing sequence types based on patterns of terms, and using recurrence relations to define sequences.

1. Proundly presents
Task 4
Alex Soh 032184 Fahmi 032799
Lailatulkadariah 033059 Nazri 032515
Nursyafiqah 032251 Shafiq 033035
Nurul Afiqah 032656 Hafizah 033006
Nurul Huda 032405 Thiba 032669

2. Topics:
4.1 Sequences
- Definition of sequence
- Example problem involving sequence.
-Types of sequence
-Give 1 example sequences use in computer programming
4.2 Mathematical Induction 1
-List down the principle of Mathematical Induction.
-Explain the method of Proof by mathematical Induction
-Give 2 example problems that use for solving mathematical
induction.

3. 4.1 Sequence

4. Definition of sequence
• A sequence is a list of numbers or a set of integers.
• In technical terms, a sequence is a function whose domain is the set of
natural numbers and whose range is a subset of the real numbers.
• We use the notation 𝑎 𝑛 to denote the image of the integer n.
• We call 𝑎 𝑛 a term of the sequence.
EXAMPLE
Consider the function 𝑎 𝑛 = 2n + 1 (explicit formulae)
The list of the terms of the sequence
𝑎1 , 𝑎2 , 𝑎3 , 𝑎4 ,𝑎5 …… (list of domain)
This function describes the sequence
3,5,7,9,11,...... (list of range)
4.1 Sequences

5. Example of problems involving sequence
The first term of an arithmetic sequence is equal to 6 and the common difference is
equal to 3. Find a formula for the n th term and the value of the 50 th term
Solution
• Use the value of the common difference d = 3 and the first term a1 = 6 in the
formula for the n th term given above
an = a1 + (n - 1 )d
= 6 + 3 (n - 1)
=3n+3
The 50 th term is found by setting n = 50 in the above formula.
a50 = 3 (50) + 3 = 153
4.1 Sequences

6. TYPES OF sequence
① ARITHMETIC PROGRESSION
Arithmetic progression is a sequence of the form
a ,a+d ,a+2d,……,a+(n-1)d, a+nd
where the initial a and the common difference d
are real numbers.
A arithmetic progression is a discrete analogue of
the linear function f(x)=dx+a.
4.1 Sequences

7. Example
• The sequences {dn} with dn= −1 + 4n and {tn} with tn= 7
− 3n are both arithmetic progressions with initial terms
and common differences equal to −1 and 4, and 7 and
−3, respectively,
If we start at n = 0. The list of terms d0, d1, d2, d3, . . .
begins with
−1, 3, 7, 11, . . . ,
and the list of terms t0, t1, t2, t3,… begins with 7, 4,
1,−2, . . . .
4.1 Sequences

8. How to find the terms
• A nth term of an arithmetic sequence can be
defined using the following formula,
an = a +(n-1)d
4.1 Sequences

9. Example
You are given that the first term of an arithmetic sequence is 1
and the 41st term is 381.
What is the 43rd term? The difference between ai and aj is d
·(j −i).
How can we use this to solve the given problem? Well since
we know a1 = 1 and a41 = 381,we have a41=381= 1+40d. So, d=
381−1 380 380
40
and a43 − a41 = 2d = 2( 40 )= 20 = 19
Therefore, a 43 = 381 + 19 = 400.
4.1 Sequences

10. • Alternatively, we can use the equation a n
=a+(n-1)d
380
a 43 = 1+ (43-1) ( )
40
380
= 1 + (42)( )
40
= 1+399
= 400
4.1 Sequences

11. Sum Of nth Terms
𝑛
• If m and n are integers m≤n the symbol 𝑘=𝑚 𝑎 𝑘 , read the
summation from k equals m to n of a-sub-k, is the sum of
all the terms am, am+1, am+2, …, an. We say that am,+ am+1+
am+2+ …+ an is the expanded form of the sum and we write
𝑛
𝑘=𝑚 𝑎 𝑘= a 𝑚+ a 𝑚 + 1+ a 𝑚 + 2+…+a 𝑛
• k= index of the summation
• m= lower limit of the summation
• n= upper limit of the summation
4.1 Sequences

12. Sum Of nth Terms (cotd’)
• If we wanted to find the sum of terms ai + ai+1
+ ai+2 + ···+ aj, we need to find the average of
the terms multiplied by the number of terms
ai+ aj
ai + ai+1 + ai+2 + ···+ aj = · (j − i + 1)
2
Note that if you every get a fractional sum from an arithmetic sequence of integers, you probably
did something wrong!
4.1 Sequences

13. Example
Let there be a sequence defined by Ai={2, 5, 8,
11, 14, 17, 20, 23}, where 0<i<9. Find the sum of
this sequence.
Using formula,
(2+23) 25
. 8−1+1 = ·8
2 2
8
𝑖=1 A i = 100.
4.1 Sequences

14. ② GEOMETRIC PROGRESSION
Geometric progression is a sequence of
form
a,ar,ar²,… arⁿ-1
where initial term a and the common ratio
r are real number, n≥0
4.1 Sequences

15. EXAMPLE
The sequences {bn}with bn = (−1)n, {cn} with cn = 2 ・ 5n, and {dn} with dn =
6 ・ (1/3)n are geometric progressions with initial term and common ratio
equal to 1 and −1; 2 and 5; and 6and 1/3, respectively.
If we start at n = 0. The list of terms b0, b1, b2, b3, b4, . . . begins with
1,−1, 1,−1, 1, . . . ;
C0,C1,C2,C3,C4
2,10,50,250,1250
d0,d1,d2,d3,d4
6,2,2 ,2 ,2 …..
3 9 24
4.1 Sequences

16. Finding nth term
• To find the nth of a GP, we must first need to find
the ratio of the GP by using the formula
𝑎𝑟 𝑛 + 1
𝑎𝑟 𝑛
Then, we can use ar 𝑛 to find the nth term of a GP
4.1 Sequences

17. EXAMPLE
• The first term of a geometric sequence of positive
integers is 1 and the 11th term is 243. How can you
find the 13th term?
g11
• We know g1= 1 and g11 = 243 so = 243 = r10. This
g1
allows us to say
gj j-i g13 12
= r , you get = r
gi g1
= (r10)6/5
= 729
4.1 Sequences

18. Summation in GP
• First formula: if a and r are real number and r≠1, then
+
𝑛 𝑟 𝑛 1−1
S n= 𝑗=0 𝑎𝑟 𝑗 𝑎( 𝑟−1 ), r≠1
𝑎
• Second formula: |r|<1 =
1−𝑟
+
𝑎−𝑟(𝑎𝑟𝑛) 𝑎𝑟 𝑛 1−𝑎
|r|≥1 = =
1−𝑟 𝑟−1
4.1 Sequences

19. EXAMPLE
• Given the term is {1, 3, 9, 27, 81}. Find the sum of
the term given.
1 − 3(81)
𝑆5 = = 121
1−3
4.1 Sequences

20. ③ HARMONIC SEQUENCE
• A harmonic sequence is a sequence h1, h2, . . . ,
1 1 1
hk such that , , …, is an arithmetic
h1 h2 hk
sequence.
4.1 Sequences

21. ④ FIBONACCI SEQUENCE
• 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, . . .
4.1 Sequences

22. 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,
namely, a0 , a1 , . . . , an-1, an ,for all integers n
with n ≥ n0 , where n0 is a nonnegative integer.
• A sequence is called a solution of a recurrence
relation if its terms satisfy the recurrence relation.
4.1 Sequences

23. • Its say that the recurrence relation is solved together
with the initial conditions when we find an explicit
formula, called a close formula, for the terms of the
sequence.
• Example: 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.
4.1 Sequences

24. EXAMPLE
• Find the Fibonacci numbers f2, f3, f4, f5, and f6.
• Solution: The recurrence relation for the Fibonacci sequence tells us that we
find successive terms by adding the previous two terms. Because the initial
conditions tell us that f0 = 0 and
f1 = 1, using the recurrence relation in the definition we find that
• f2 = f1 + f0 = 1 + 0 = 1,
• f3 = f2 + f1 = 1 + 1 = 2,
• f4 = f3 + f2 = 2 + 1 = 3,
• f5 = f4 + f3 = 3 + 2 = 5,
• f6 = f5 + f4 = 5 + 3 = 8.
4.1 Sequences

25. ⑤ Special Integer Sequences
• It is used to identify a sequence
• Example: Find the formulae for the sequences
with the following first 5 terms
a) 1, 1/2, 1/4, 1/8, 1/16
b) 1,3,5,7,9
4.1 Sequences

26. • Solution:
a) we recognize that the denominator are powers of 2.
the sequence with an = (1/2)n , n=0,1,2,... is a
possible match. This proposed sequence is a
geometric progression with a=1 and r= 1/2 .
b) 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. This proposed
sequence is an arithmetic progression with a=1
and d=2.
4.1 Sequences

27. 4.1 Sequences

28. 4.1 Sequences

29. Index shifting
• Sometimes we shift the index of summation in a sum. This is often
done when two sums need to be added but their indices of
summation do not match.
• Example: we have 5 𝑗2 but we want the index of summation to
𝑗=1
run between 0 and 4 rather than 1 to 5. Then we let k=j-1=0 and 𝑗2
become (k+1)2 . Hence,
5
𝑗=1 𝑗2= 4 (𝑘
𝑘=0 + 1)2
=1 + 4 + 9 + 16 + 25 = 55
4.1 Sequences

30. Double summation
4 3 4
𝑖=1 𝑗=1 𝑖𝑗 = 𝑖=1(𝑖 + 2𝑖 + 3𝑖)
4
= 𝑖=1 6𝑖
=6+12+18+24
=60
4.1 Sequences

31. usage on computer programming
• A structured series of shots or scenes with a
beginning, middle and end , the term sequence
can be applied to video, audio or graphics.
• Structured programming provides a number of
constructs that are used to define the sequence
in which the program statements are to be
executed.

32. Usage on computer programming
FLOWCHART
1
2
3
4.1 Sequences

33. Pseudocode
Statement-1
Statement-2
Statement-3
Example:
input a
b= 5 + 2 * a
print b
4.1 Sequences

34. 4.2 Mathematical
Induction

35. Principle of Mathematical Induction
To prove that P(n) is true for all positive intergers n, where P(n) is a propositional
function by 2 steps:
BASIS: we verify that P(1) is true @ show that a initial value is true for all Z+ of the
propositional function (inductive hypothesis )
INDUCTIVE: we show that the conditional statement ∀k (P(k) → P(k+1)) is true for all
Z+ of k.
4.2 Mathematical Induction 1

36. • Similarly, we can say that mathematical
induction is a method for proving a property
defined that the property for integer n is true
for all values of n that are greater than or
equal to some initial interger.
P(1)^∀k(P(k) → P(k+1))) → ∀nP(n)
4.2 Mathematical Induction 1

37. Method of proof
• The proofs of the basis and inductive steps shown in the example
illustrate 2 different ways to show an equation is true
Transforming LHS and RHS independently until they
seem to be equal.
Transforming one side of equation until it is seen to
be the same as the other side of the equation.
4.2 Mathematical Induction 1

38. Problem Solving
• Example 1
Using Mathematical Induction, prove that
𝑛(𝑛+1)
1+2+…+n = 2
, for all integers n≥1
4.2 Mathematical Induction 1

39. Solution: we know that the property P(n) is the equation shown at the previous
page. Hence, we need to prove it using BASIS and INDUCTION steps.
BASIS: Show that P(1) is true for the LHS and RHS of the equation.
𝑛(𝑛+1)
LHS= 1+2+…+n RHS= , n≥1
=1 2
1(1+1)
= 2
2
= 2
=1
∴Since LHS = RHS, we have proven P(1) is true.

40. INDUCTIVE: we need to show that the equation can take any value k and it’s successive
value (k+1) by defining P(k) [the inductive hypothesis] and P(k+1) for k≥1. If P(k) is true
then P(k+1) is true.
Inductive Hypothesis P(k+1)= 1+2+…+k+(k+1) P(k+1)= 1+2+…+k+(k+1)
P(k)= 1+2+…+k (𝑘+1)(𝑘+1+1) = P(k)+ (k+1)
=
𝑘(𝑘+1) 2 𝑘2+𝑘
= 2 (𝑘+1)(𝑘+2) = 2 + (k+1)
=
𝑘2+𝑘 2
2 𝑘2+𝑘 +2(𝑘+1)
= 2 𝑘 +3𝑘+2 =
= 2
-Eq.1 2
2
(assume that it is true) 𝑘 +𝑘+2𝑘+2
= 2
𝑘2+3𝑘+2
= 2
-Eq.2
∴Since Eq.1 and Eq.2 is identical for both LHS and RHS, therefore P(k+1) is true.

41. • Example 2
Show that the sum of the first n odd integers
is n2
Example: If n = 5, 1+3+5+7+9 = 25 = 52
Here, we know that d= n2-n1 =2 and a=1
Hence,
an = 1+(n-1)2 𝑛
Sn = 𝑖=0(2𝑛 − 1)
= 1+ (n)2 𝑛
P(n)= 𝑖=0(2𝑛 − 1)
= 1+2n-2 𝑛
n2 = 𝑖=0(2𝑛 − 1)
=2n-1

42. Solution: Again, we have defined the general formula for the sum of the nth term of
the odd positive integers as P(n)
BASIS: Show that P(1) is true for the LHS and RHS of the equation.
LHS RHS
𝑛
n 2= 1 2 P(n) = 𝑖=1(2𝑛 − 1) , n≥1
=1 P(1) = 1 (2(1) − 1)
𝑖=1
=1
∴Since LHS = RHS, we have proven P(1) is true.

43. INDUCTIVE: Again we need to show that if P(k) is true then P(k+1) is true.
P(k+1)= 1+3+…+(2k)
𝑘+1
= 𝑖=0 (2𝑖 − 1)
Inductive Hypothesis = 𝑘+1 2 (LHS)
P(k)= 1+3+…+(2k-1)
𝑘 𝑘+1
= 𝑖=1(2𝑖 − 1) 𝑖=1 (2𝑖 − 1)= 𝑘 + 1 2
= 𝑘2 2(k+1)-1+ 𝑘
𝑖=1(2𝑖 − 1) = 𝑘 + 1 2
(assume that it is true) 2k+1+k2= 𝑘 + 1 2
k2+2k+1= k2 +2k+1 (RHS)
∴Since LHS = RHS, therefore P(k+1) is true.