Chapter 5- Sequences and Mathematical Induction (I)

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.