Mathematics

1. Mathematics

2. 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=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.

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. Ex:1.3+2.32+3.33+......+n.3n= 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 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. Show by PMI that 1.3+2.32+3.33+......+n.3n= Illustrative Example 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 1.3+2.32+3.33+......+n.3n= (Proved)

15. 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. (m+1)3+5(m+1)=(m3+5m)+(3m2+3m+6) = 3k´ = 3k+3(m2+m+2) ( m2 + m + 2 I ) 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’

16. (m+1)3+5(m+1)=3k’ P(m+1) is divisible by 3  P(m)  P (m+1) P(n) is true for all n N Step4:  n3+5n is divisible by ‘3’ for n N Solution Continued

17. Prove that mathematical induction that 72n + 3n – 1(23n – 3) is divisible by 25, . Class Exercise - 6 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.

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 + 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

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) =n(n2 -1+6) =n(n2-1)+6n =n(n-1)(n+1)+6n 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. Prove by PMI that • 1.2.3. + 2.3.4 + 3.4.5 + ... + • n(n + 1) (n + 2) = step1. P(1): LHS=1.2.3=6  L.H.S=R.H.S Step2. Assume P(m) is true Class Exercise -3 Solution:

23. 1.2.3.+2.3.4+...+m(m+1)(m+2)+(m+1)(m+2)(m+3)= Solution Continued Adding (m+1)(m+2)(m+3)

24. 1.2.3.+2.3.4+...+(m+1)(m+2)(m+3) Solution Continued Step4. As P(m)  P (m+1) P(n) is true for all n N •  1.2.3.+2.3.4+3.4.5+...+n(n+1)(n+2)=

25. Let P(n) : an + bn = cn + dn For n = 1, P(1) : a+b=c+d n = 2, P(2) : a2+b2 = c2+d2 Assume P(m) and P(m + 1) hold true • am+bm = cm+dm • am+1+bm+1= cm+1+dm+1 Class Exercise -4 • If a+b=c+d and a2+b2=c2+d2 , then show by mathematical induction, an+bn = cn+dn Solution: P(1) and P(2) hold true.

26. 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 Solution Continued

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

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

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

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

31. For n = 1, LHS =7 • RHS = 7 • LHS = RHS P(m):7+77+777+...+77...7(m times) Class Exercise -9 Solution: Let P(n) holds true for n = m

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

33. 7+77+...+77...7(m + 1) times  P(m + 1) holds true  P(n) is true 7+77+777+...+77...7(n times) Solution Continued

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

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

36. Thank you