Alex Soh 032184 Fahmi 032799

Lailatulkadariah 033059 Nazri 032515

Nursyafiqah 032251 Shafiq 033035

Nurul Afiqah 032656 Hafizah 033006

Nurul Huda 032405 Thiba 032669

Proundly presents Task 4

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.

4.1 Sequence

4.1 Sequences

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 𝑎𝑛to denote the image of the integer n. • We call 𝑎𝑛 a term of the sequence.

EXAMPLE Consider the function 𝑎𝑛 = 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)

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) = 3 n + 3

The 50 th term is found by setting n = 50 in the above formula.

a50 = 3 (50) + 3 = 153

4.1 Sequences

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

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

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

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

40 and a43 − a41 = 2d = 2(

380

40)=

380

20 = 19

Therefore, a 43 = 381 + 19 = 400.

4.1 Sequences

• Alternatively, we can use the equation a n =a+(n-1)d

a 43 = 1+ (43-1) (380

40)

= 1 + (42)(380

40)

= 1+399

= 400

4.1 Sequences

Sum Of nth Terms • If m and n are integers m≤n the symbol 𝑎𝑘

𝑛𝑘=𝑚 , 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

𝑎𝑘𝑛𝑘=𝑚 = a𝑚+ a𝑚 + 1+ a𝑚 + 2+…+a𝑛

• k= index of the summation • m= lower limit of the summation • n= upper limit of the summation

4.1 Sequences

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 + ai+1 + ai+2 + · · · + aj = ai+ aj

2· (j − i + 1)

Note that if you every get a fractional sum from an arithmetic sequence of integers, you probably did something wrong!

4.1 Sequences

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)

2. 8 − 1 + 1 =

25

2· 8

A i

8𝑖=1 = 100.

4.1 Sequences

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

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

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

𝑎𝑟𝑛 + 1

𝑎𝑟𝑛

Then, we can use ar𝑛 to find the nth term of a GP

4.1 Sequences

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?

• We know g1= 1 and g11 = 243 so g11 g1

= 243 = r10. This

allows us to say

gj

gi= rj-i, you get

g13

g1

= r12

= (r10)6/5 = 729

4.1 Sequences

Summation in GP

• First formula: if a and r are real number and r≠1, then

Sn= 𝑎𝑟𝑗𝑛𝑗=0 𝑎(

𝑟𝑛+1−1

𝑟−1), r≠1

• Second formula: |r|<1 = 𝑎

1−𝑟

|r|≥1 = 𝑎−𝑟(𝑎𝑟𝑛)

1−𝑟 =

𝑎𝑟𝑛+1−𝑎

𝑟−1

4.1 Sequences

EXAMPLE

• Given the term is {1, 3, 9, 27, 81}. Find the sum of the term given.

𝑆5 = 1 − 3(81)

1 − 3= 121

4.1 Sequences

③ HARMONIC SEQUENCE

• A harmonic sequence is a sequence h1, h2, . . . ,

hk such that1h1

, 1h2

, …, 1hk

is an arithmetic

sequence.

4.1 Sequences

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

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

• 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

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

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

• 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

4.1 Sequences

4.1 Sequences

• 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 𝑗25𝑗=1 but we want the index of summation to

run between 0 and 4 rather than 1 to 5. Then we let k=j-1=0 and 𝑗2 become (k+1)2 . Hence,

𝑗25𝑗=1 = (𝑘 + 1)24

𝑘=0

=1 + 4 + 9 + 16 + 25 = 55

Index shifting

4.1 Sequences

𝑖𝑗3𝑗=1

4𝑖=1 = (𝑖 + 2𝑖 + 3𝑖)4

𝑖=1

= 6𝑖4𝑖=1

=6+12+18+24

=60

Double summation

4.1 Sequences

• 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.

usage on computer programming

FLOWCHART

1

2

3

Usage on computer programming

4.1 Sequences

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

4.1 Sequences

4.2 Mathematical

Induction

To prove that P(n) is true for all positive intergers n, where P(n) is a propositional function by 2 steps:

4.2 Mathematical Induction 1

Principle of Mathematical Induction

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.

• 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

• 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.

Method of proof

4.2 Mathematical Induction 1

• Example 1

Using Mathematical Induction, prove that

4.2 Mathematical Induction 1

Problem Solving

1+2+…+n = 𝑛(𝑛+1)

2, for all integers n≥1

BASIS: Show that P(1) is true for the LHS and RHS of the equation. ∴Since LHS = RHS, we have proven P(1) is true.

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.

RHS= 𝑛(𝑛+1)

2 , n≥1

= 1(1+1)

2

= 2

2

=1

LHS= 1+2+…+n = 1

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.

∴Since Eq.1 and Eq.2 is identical for both LHS and RHS, therefore P(k+1) is true.

Inductive Hypothesis P(k)= 1+2+…+k

= 𝑘(𝑘+1)

2

= 𝑘2+𝑘

2

(assume that it is true)

P(k+1)= 1+2+…+k+(k+1)

= (𝑘+1)(𝑘+1+1)

2

= (𝑘+1)(𝑘+2)

2

= 𝑘2+3𝑘+2

2 -Eq.1

P(k+1)= 1+2+…+k+(k+1) = P(k)+ (k+1)

= 𝑘2+𝑘

2+ (k+1)

=𝑘2+𝑘 +2(𝑘+1)

2

= 𝑘2+𝑘+2𝑘+2

2

= 𝑘2+3𝑘+2

2 -Eq.2

• 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 = 1+ (n)2 = 1+2n-2

=2n-1

Sn = (2𝑛 − 1)𝑛𝑖=0

P(n)= (2𝑛 − 1)𝑛𝑖=0

n2 = (2𝑛 − 1)𝑛𝑖=0

BASIS: Show that P(1) is true for the LHS and RHS of the equation. ∴Since LHS = RHS, we have proven P(1) is true.

Solution: Again, we have defined the general formula for the sum of the nth term of the odd positive integers as P(n)

RHS P(n) = (2𝑛 − 1)𝑛

𝑖=1 , n≥1 P(1) = (2(1) − 1)1

𝑖=1 = 1

LHS

n2= 12 = 1

INDUCTIVE: Again we need to show that if P(k) is true then P(k+1) is true.

∴Since LHS = RHS, therefore P(k+1) is true.

Inductive Hypothesis P(k)= 1+3+…+(2k-1)

= (2𝑖 − 1)𝑘𝑖=1

= 𝑘2

(assume that it is true)

P(k+1)= 1+3+…+(2k) = (2𝑖 − 1)𝑘+1

𝑖=0

= 𝑘 + 1 2 (LHS)

(2𝑖 − 1)𝑘+1𝑖=1 = 𝑘 + 1 2

2(k+1)-1+ (2𝑖 − 1)𝑘𝑖=1 = 𝑘 + 1 2

2k+1+k2= 𝑘 + 1 2 k2+2k+1= k2 +2k+1 (RHS)