Mathematics. Principle of Mathematical Induction Session.

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