Mathematics

Principle of Mathematical Induction

Session

Session Objectives

Session Objective

1. Introduction

2. Steps involved in the use of mathematical induction

3. Principle of mathematical induction.

Statement

Statement:- A sentence which can be judged as true or false.

Example:1. ‘2 is only even prime number’ 2. ‘Bagdad is capital of Iraq’ 3. ‘2n+5 is always divisible by 5 for all nN

Mathematical statement:

Example 1 and 3.

Induction

Induction :It’s a process Particular General

Example: Statement- ’2n+1’ is odd number.

n=1 2.1+1=3 is odd. True

n=2 2.2+1=5 is odd. True

n=3 2.3+1=7 is odd. True

Observation tentative conclusion (‘2n+1 is odd’)

let its true for n=m. i.e 2m+1 is odd.

Induction

for n=m+1

2(m+1)+1 =2m+1+2

odd +2=odd

’2n+1 is odd for all n’

Now it is Generalized

Induction

Steps Involved: 1. Verification2. Induction3. Generalization.

Important: Process of Mathematical Induction (PMI) is applicable for natural numbers.

Usage: 1. to prove mathematical formula

2. to check divisibility of a expression by a number

Ex:1.3+2.32+3.33+......+n.3n= n 1(2n 1)3 3

4

Ex: Prove n3+5n is divisible by ‘3’.

Algorithm

Let P(n) be the given statement.

Step 1: Prove P(1) is true Verification

Step 2: Assume P(n) is true for some n=mN i.e. P(m) is true.

Step 3: Using above assumption prove P(m+1) is true. i.e P(m) P (m+1)

Step 4: Above steps lead us to generalize the fact P(n) is true for all n N.

Questions

Illustrative Example

Principle of mathematical induction is applicable to

(a) set of integers

(b) set of real numbers

(c) set of positive integers

(d) None of these

Solution : (c)

Principle of mathematical induction is applicable to natural numbers or set of positive integers only.

Illustrative Example

Show by PMI that 1.3+2.32+3.33+......+n.3n=

n 1(2n 1)3 3

4Solution:

Step 1. for n=1, p(1)=1. 3=3 (LHS)

n 1 2(2n 1)3 3 (2 1 1)3 3R.H.S 3

4 4

L.H.S=R.H.S P(1) is true

Step2. Assume that P(m) is true

m 12 m (2m 1)3 3

P(m) : 1 .3 + 2 . 3 + .... + m .3 =4

Solution Continued

2 m

m 1

P(m) : 1.3 + 2.3 + .... + m.3

(2m 1)3 3=

4

Step3: To prove P(m + 1) holds true Adding. (m + 1).3m+1 to both sides

m 1 m 12 m m 1 (2m 1)3 3 4(m 1)3

1.3 2.3 ... m.3 m 1 .34

m 1 m 1(2m 1) 3 4(m 1)3 3

4

m 13 (2m 1 4m 4) 3

4

m 1(6m 3)3 3

4

m 1 m 23(2m 1) 3 3 (2m 1)3 3

4 4P(m) P (m+1)

Solution Continued

Step4. As P(m) P (m+1) P(n) is true for all n N

1.3+2.32+3.33+......+n.3n= n 1(2n 1)3 3

4

(Proved)

2 m m 1

m 2

1.3 2.3 ... m.3 m 1 .3

(2m 1)3 3

4

(m 1) 1{2(m 1) 1}3 3

4

Illustrative Example

Prove n3+5n is divisible by ‘3’ for n N (By PMI or Otherwise)

Solution: P(n) : ‘n3+5n is divisible by 3’

Step1: P(1) = ‘6 divisible by 3’ which is true

Step2: For some n=m, P(m) holds truei.e. m3+5m=3k, k N

step3: To prove P(m+1) holds true, we have to prove that (m+1)3+5(m+1) is divisible by 3.

= 3k´

(m+1)3+5(m+1)=(m3+5m)+(3m2+3m+6)

= 3k+3(m2+m+2)

( m2 + m + 2 I )

Solution Continued

(m+1)3+5(m+1)=3k’

P(m+1) is divisible by 3

P(m) P (m+1)

Step4: P(n) is true for all n N

n3+5n is divisible by ‘3’ for n N

Class Exercise - 6

Prove that mathematical induction that 72n + 3n – 1(23n – 3) is divisible by 25, .n N

Solution :

Let P(n) : 72n + (23n – 3)3n – 1 is divisible by 25.

Step I: n = 1

P(1) = 72 + (23 – 3)31 – 1

= 72 + 20 · 30

= 49 + 1 = 50 As P(1) is 50 which is divisible by 25, hence P(1) is true.

Solution Continued

Step II: Assuming P(m) is divisible by 25,

P(m) = 72m + (23m – 3)3m – 1 = 25(K) ... (i)

(K is a positive integer.)

Now P(m + 1) = 72(m + 1) + (23(m + 1) – 3)3m + 1 – 1

= 72m + 2 + (23m + 3 – 3)3m + 1 – 1

= 49 × 72m + 8(23m – 3)3m – 1 × 3

= 49 × 72m + 24(23m – 3)3m – 1

With the help of equation (i), we can write the above expression as

Solution Continued

3m 3 m 1 3m 3 m 149 25K 2 3 24 2 3

3m 3 m 1 3m 3 m 149 25K 49 2 3 24 2 3

3m 3 m 149 25K 2 3 49 24 3m 3 m 149 25K 25 2 3

3m 3 m 125 49k 2 3

3m 3 m 1P m 1 25 49k 2 3

Now from the above equation, we can conclude that P(m + 1) is divisible by 25.

Hence, P(n) is divisible by 25 for all natural numbers.

Alternative Solution

Alternative Method: without PMI

n3+5n = n(n2+5)

=n(n2 -1+6)

=n(n2-1)+6n

=n(n-1)(n+1)+6n

Product of three consecutive numbers

Illustrative Example

P(n) is the statement ‘n2 – n + 41 is prime’Verify it.

Solution:

For n = 1 P(1) = ‘41 is a prime’ True.

For n = 2 P(2) = ‘43 is a prime’ True.

But for n = 41 P(41) = ‘412 is a prime’ False.

False Statement

Class Exercise -3Prove by PMI that

1.2.3. + 2.3.4 + 3.4.5 + ... +

n(n + 1) (n + 2) = n(n 1) n 2 n 3

4

Solution:

step1. P(1): LHS=1.2.3=6

1.(1 1).(1 2).(1 3)R.H.S 6

4 L.H.S=R.H.S

Step2. Assume P(m) is true

m(m 1) m 2 m 31.2.3.+2.3.4+..+m(m+1)(m+ 2) =

4

Solution Continued

P(m): 1.2.3.+2.3.4+..+m(m+1)(m+ 2)

m(m 1) m 2 m 3=

4

Adding (m+1)(m+2)(m+3)

1.2.3.+2.3.4+...+m(m+1)(m+2)+(m+1)(m+2)(m+3)=

m(m 1) m 2 m 3(m 1) m 2 m 3

4

m(m 1) m 2 m 3 4(m 1) m 2 m 3

4

(m 1) m 2 m 3 (m 4)

4

Solution Continued

Step4. As P(m) P (m+1) P(n) is true for all n N

1.2.3.+2.3.4+...+(m+1)(m+2)(m+3)

(m 1) m 2 m 3 (m 4)

4

1.2.3.+2.3.4+3.4.5+...+n(n+1)(n+2)=

n(n 1) n 2 n 3

4

Class Exercise -4

If a+b=c+d and a2+b2=c2+d2 , then show by mathematical induction, an+bn = cn+dn

Solution: Let P(n) : an + bn = cn + dn

n = 2, P(2) : a2+b2 = c2+d2

For n = 1, P(1) : a+b=c+d

P(1) and P(2) hold true.

Assume P(m) and P(m + 1) hold true

am+bm = cm+dm

am+1+bm+1= cm+1+dm+1

Solution Continued

a+b = c+d; a2+b2=c2+d2

am+bm = cm+dm; am+1+bm+1= cm+1+dm+1

P(m+2) : am+2+bm+2

= (a+b)(am+1+bm+1)–ab(am+bm)

= (c+d)(cm+1+dm+1)–cd(cm+dm)

= cm + 2 + dm + 2

P(m+2) holds true.

an+bn = cn+dn holds true for n N

Class Exercise - 7

nn 1

n

UsingPMI prove that

sin2cos cos2 cos 4 ... cos2

2 sin

Solution:

nn 1

n

sin2Let P(n) : cos cos2 cos 4 ... cos2

2 sin

For n = 1, LHS = Cos

sin2R.H.S cos

2sin

Assume P(m) holds true

mm 1

m

sin2cos .cos2 .cos 4 ... cos2

2 sin

Solution Continued

mm 1

m

sin2cos .cos2 .cos 4 ... cos2

2 sin

multiplying both sides by Cos2m m 1 mcos cos2 cos 4 ...cos2 cos2

m m

m

sin2 .cos2

2 sin

2m m

m 1

2 sin cos2

2 sin

m 1

m 1

sin2

2 sin :P(m+1) holds true

As P(m) P(m+1) P(n) is true for all n N

nn 1

n

sin2cos .cos2 .cos 4 ... cos2

2 sin

Class Exercise - 8

21Prove that 1 + 2 + 3 + ... + n< (2n + 1)

8

Solution:

For n = 1, LHS = 1;RHS =9/8 LHS < RHS

Let assume P(m) in true

211 + 2 + 3 + ... + m< (2m + 1)

8

211 + 2 + 3 + ... + m+(m+1)< (2m + 1) (m 1)

8

24m 4m 1

m 18

Solution Continued

24m 4m 11 + 2 + 3 + ... + m+(m+1)< m 1

8

24m 12m 9

8

211 + 2 + 3 + ... + m+(m+1)< (2m 3)

8

211 + 2 + 3 + ... + m+(m+1)< {2(m 1) 1}

8

P (m+1) holds true

P(n) holds true n N

Class Exercise -9

n 1

Prove by PMI that

7 + 77 + 777 + ... + 777... 7 (n times)

7 = 10 9n 10

81

Solution:

For n = 1, LHS =7

RHS = 7

LHS = RHS

2710 9 10

81

Let P(n) holds true for n = m

P(m):7+77+777+...+77...7(m times) m 1710 9m 10

81

Solution Continued

7+77+777+...+77...7(m times)

m 1710 9m 10

81

Adding77...7(m+1)times to both sides

7+77+...+77...7(m times)+77...7(m + 1) times

m 1710 9m 10 + 777 ... 7 (m + 1) times

81

m 17 7= 10 9m 10 99...9(m 1)times

81 9

m 1 m 17 710 9m 10 10 1

81 9

Solution Continued

7+77+...+77...7(m + 1) times

m 1 m 17 710 9m 10 10 1

81 9

m 1 m 17

10 9m 10 9.10 981

m 17

10.10 9 m 1 1081

m 27

10 9 m 1 1081

P(m + 1) holds true P(n) is true

7+77+777+...+77...7(n times) n 1710 9m 10 n N

81

Class Exercise -10

2

1 1 1 1 11 ... 2 n 1; n N

4 9 16 nn

Prove that

Solution :- 2

1 1 1 1Let P(n) : 1 ... 2 n 1

4 9 nn

For n = 2, 1 5

LHS 14 4

1 3 6

RHS 22 2 4

LHS RHS

P(n) is true for n 2

2

1 1 1 1Let P(m) : 1 ... 2 for m 1....(1)

4 9 mm

2

1Now Adding to both side in equ. (1)

m 1

Solution Continued

2 2

1 1 1 1We get 1 ...

4 9 m m 1

2

1 12

m m 1

2

1 12

m m 1

2

2

m 2m 1 m2

2 m 1

2

m(m 1) 12

m m 1

2

1 1 12 2

m 1 m 1m m 1

2

10 as m N

m m 1 P(m + 1) holds true.

P(n) holds true , n > 1. n N

Thank you