1. Mathematics

2. Principle of Mathematical Induction
Session
Principle of Mathematical Induction

3. Session Objectives

4. Session Objective 1. Introduction
2. Steps involved in the use of mathematical induction
3. Principle of mathematical induction.

5. 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 nN
Mathematical statement:
Example 1 and 3.

6. Induction Induction :It’s a process Particular  General
Example: Statement- ’2n+1’ is odd number.
n= =3 is odd.
True
n= =5 is odd.
True
n= =7 is odd.
True
Observation  tentative conclusion (‘2n+1 is odd’)
let its true for n=m. i.e 2m+1 is odd.

7. Induction for n=m+1 2(m+1)+1 =2m+1+2 odd +2=odd Now it is Generalized
’2n+1 is odd for all n’

8. Induction Steps Involved: 1. Verification 2. Induction
3. Generalization.
Important: Process of Mathematical Induction (PMI) is applicable for natural numbers.
Usage: 1. to prove mathematical formula
Ex: n.3n=
2. to check divisibility of a expression by a number
Ex: Prove n3+5n is divisible by ‘3’.

9. 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=mN
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.

10. Questions

11. Illustrative Example
Principle of mathematical induction is applicable to
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.

12. Illustrative Example Show by PMI that 1.3+2.32+3.33+......+n.3n=
Solution:
Step 1.
for n=1, p(1)=1. 3=3 (LHS)
L.H.S=R.H.S
P(1) is true
Step2. Assume that P(m) is true

13. Solution Continued Step3: To prove P(m + 1) holds true
Adding. (m + 1).3m+1 to both sides
P(m)  P (m+1)

14. Solution Continued Step4. As P(m)  P (m+1) P(n) is true for all n N
(Proved)

15. 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 true
i.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 )

16. 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

17. Class Exercise - 6
Prove that mathematical induction that 72n + 3n – 1(23n – 3) is divisible by 25,
Solution :
Let P(n) : 72n + (23n – 3)3n – 1 is divisible by 25.
Step I: n = 1
P(1) = 72 + (23 – 3)31 – 1
= · 30
= = 50 As P(1) is 50 which is divisible by 25, hence P(1) is true.

18. 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 (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

19. Solution Continued
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.

20. Alternative Solution Alternative Method: without PMI n3+5n = n(n2+5)
Product of three
consecutive numbers

21. 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

22. Class Exercise -3 Prove by PMI that 1.2.3. + 2.3.4 + 3.4.5 + ... +
n(n + 1) (n + 2) =
Solution:
step1.
P(1): LHS=1.2.3=6
 L.H.S=R.H.S
Step2. Assume P(m) is true

23. Solution Continued Adding (m+1)(m+2)(m+3)
m(m+1)(m+2)+(m+1)(m+2)(m+3)=

24. Solution Continued 1.2.3.+2.3.4+...+(m+1)(m+2)(m+3) Step4.
As P(m)  P (m+1)
P(n) is true for all n N
 n(n+1)(n+2)=

25. 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

26. 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 dm + 2
 P(m+2) holds true.
an+bn = cn+dn holds true for n N

27. Class Exercise - 7 Solution: For n = 1, LHS = Cos
Assume P(m) holds true

28. Solution Continued multiplying both sides by Cos2m :P(m+1) holds true
As P(m)  P(m+1) P(n) is true for all n N

29. Class Exercise - 8 Solution: For n = 1, LHS = 1;RHS =9/8 LHS < RHS
Let assume P(m) in true

30. Solution Continued
P (m+1) holds true
 P(n) holds true  n N

31. Class Exercise -9 Solution: For n = 1, LHS =7 RHS = 7 LHS = RHS
Let P(n) holds true for n = m
P(m): (m times)

32. Solution Continued 7+77+777+...+77...7(m times)
Adding77...7(m+1)times to both sides
(m times) (m + 1) times

33. Solution Continued 7+77+...+77...7(m + 1) times
 P(m + 1) holds true  P(n) is true
 (n times)

34. Class Exercise -10
Prove that
Solution :-
For n = 2,

35. Solution Continued
P(m + 1) holds true.
P(n) holds true , n > 1.

36. Thank you