# Important Questions for CBSE Class 11 Maths Chapter 4 - Principle of Mathematical Induction

## CBSE Class 11 Maths Chapter-4 Important Questions - Free PDF Download

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4 Marks Questions

1. For every integer n, prove that

Ans. P(n) : 7n­­­­-3n is divisible by 4

For n=1

P(1) : 71 -31 =4 which is divisible by Thus, P(1) is true

Let P(k) be true

2. Prove that

Ans.

3. Prove that

Ans.

4. Prove

Ans.

5. Prove 1.2+2.3+3.4+--+n

Ans.

6. Prove (2n+7)<(n+3)2

Ans.

7. Prove

Ans.

8. Prove 1.2+2.22+3.23+…+n.2n = (n-1)2 n + 1+2

Ans.

c

9. Prove that 2.7n + 3.5n – 5 is divisible by 24

Ans. P(n) : 2.7n + 3.5n – 5 is divisible by 24

for n = 1

P (1) : 2.71 + 3.51 – 5 = 24 is divisible by 24

Hence result is true for n = 1

Let P (K) be true

P (K) : 2.7K + 3.5K – 5

(i)

we want to prove that P (K+!) is True whenever P (K) is true

2.7K+1 + 3.5K+1 – 5 = 2.7K .71 + 3.5K . 51 -5

Hence by P M I p (n) is true for all n N.

10. Prove that 41n – 14n is a multiple of 27

Ans. P (n) : 41n – 14n is a multiple of 27

for n = 1

P (1) : 411 – 14 = 27, which is a multiple of 27

Let P (K) be True

P (K) : 41K – 14K

(i)

we want to prove that result is true for n = K + 1

41K+1– 14k+1 = 41K.41-14K. 14

is a multiple of 27

Hence by PMI p (n) is true for ace n N.

11. Using induction, prove that 10n + 3.4n+2 + 5 is divisible by 9 .

Ans. P (n) : 10n + 3.4n+2 + 5 is divisible by 9

For n = 1

p (1) : 101 + 3.41+2 + 5=207, divisible by 9

Hence result is true for n = 1

Let p (K) be true

p (K) : 10K +3.4K+2 + 5 is divisible by 9

we want to prove that result is true for n = K + 1

10(K+1) +3.4K+1+2+5=10K+1+3.4K+3+5

which is divisible by 9.

12. Prove that 12 + 22 + 32 + ---+ n2 =

Ans. Let P(n) : 12 + 22 + 32 + -- +n2 =

for n = 1

Thus P (K+1) is true, whenever P (K) is true.

Hence, from PMI, the statement P (n) is true for all natural no. n.

13. Prove that 1+3+32 + -- + 3n-1 =

Ans. Let

P (n) : 1 + 3 + 32 + -- + 3n-1 = for n = 1

P (1) 31-1 =

which is true

Let P (K) be true

P (K) : 1 + 3 + 32 + ---- + 3K-1 =

we want to prove that P (K+1) is true

Hence p (K+1) is true whenever p (K) is True

14. By induction, prove that 12 + 22 + 32 + -- + n2 >

Ans. Let P (n) : 12 + 22 + 32 + --- + n2 >

for n = 1

12 > which is true

Let P (K) be true

P (K) : 12 + 22 + 32 + -- + K2 >

we want to prove that P (K + 1) is true

P (K+1) : 12 + 22 + --- + (K+1)2

Hence by PMI P (n) is true

15. Prove by PMI (ab)n = an bn

Ans. Let P (n) : (ab)n = an bn

for n = 1

ab = ab which is true

Let P (K) be true

(ab)K = aK bK (1)

we want to prove that P (K+1) is true

(ab)K+1 = aK+1. bK+1

L.H.S = (ab)K+1

16. Prove by PMI a + ar + ar2 + ---- + arn-1 =

Ans. Let P (n) : a + ar + ar2 + -- + arn-1 =

for n = 1

P (1) = a = a which is true

Let P (K) be true

P (K) : a + ar + ar2 + -- + arK-1 =

we want to prove that

P (K+1) : a + ar + ar2 + -- + arK =

L.H.S

Thus P (K+1) is true whenever P(K) is true

Hence by PMI P(n) is true for all n N

17. Prove that x2n – y2n is divisible by x + y.

Ans. P (n) : x2n – y2n is divisible by x + y

for n = 1

p (1) : x2 – y2 = (x – y) (x + y), which is divisible by x + y

Hence result is true for n = 1

Let P (K) be true

p (K) : x2K – y2K is divisible by x + y

18. Prove that n (n + 1) (2n + 1) is divisible by 6.

Ans.P (n) : n (n+1) (2n+1) is divisible by 6 for n = 1

P (1) : (1) (2) (3) = 6 is divisible by 6

Hence result is true for n = 1

Let P (K) be true

P (K) : K (K+1) (2K+1) is divisible by 6

19. Show that 23n – 1 is divisible by 7 .

Ans.P (n) : 23n – 1 is divisible by 7

for n = 1

P (1) : 23 – 1 = 7 which is divisible by 7

Let P (K) be true

P (K) : 23K – 1 is divisible by 7

s

Thus P (K+1) is true

Hence by P.M.I P (n) is true

20. Prove by P M I.

1. 2. 3 + 2. 3. 4 + --- + n (n + 1) (n + 2)

=

Ans.Let P (n) : 1. 2. 3 + 2. 3. 4 + --- + n (n+1) (n+2)

=

For n = 1

P (1) = 1 (2) (3) =

P (1) = 6 = 6 which is true

Let P (K) be true

P (K) : 1. 2. 3 + 2. 3. 4 + -- + K (K+1) (K+2)

=

we want to prove that

P (K+1) n: 1. 2. 3 + 2. 3. 4 + -- + (K+1) (K+2) (K+3) =

L.H.S = 1. 2. 3 + 2. 3. 4 + -- + K (K+1) (K+2) + (K+1) (K+2) (K+3)

Thus P (K+1) is true whenever P(K) is true.

21. Prove that

Ans.P(n) :

For n = 1

P (1) =

Let P (K) be true

Thus P (K+1) is for whenever P (K) is true.

22. Show that the sum of the first n odd natural no is n2.

Ans.Let P (n) : 1 + 3 + 5 + --- + (2n-1) = n2

For n = 1

P (1) = 1 = 1 which is true

Let P (K) be true

P (K) : 1 + 3 + 5 + --- + (2K-1) = K2 (1)

we want to prove that P (K+1) is true

P (K+1) : 1 + 3 + 5 + --- + (2K+1) = (K+1)2

L.H.S =

Thus P (K+1) is true whenever P(K) is true.

Hence by PMI, P(n) is true for all n N.

23. Prove by P M I

Ans.P (n) :

For n = 1

P (1) : 13 = 13 which is true

Let P (K) be true

Thus P (K+1) is true whenever P (K) is true.

24. Prove.

Ans.P (n) :

For n = 1

P (1) : 4 = 4 which is true

Let P (K) be true

P (K) :

We want to power that P (K+1) is true

Thus P (K+1) is true whenever P (K) is true.

25. Prove that 32n+2 – 8n – 9 is divisible by 8

Ans.P(n) : 32n+2 – 8n – 9 is divisible by 8

For n = 1

P (1) : 32+2 – 8 – 9 = 64

which is divisible by 8

Hence result is true for n = 1

Let P (K) be true

P (K) : 32K+2 – 8K – 9 is divisible by 8

( from i)

26. Prove by PMI.

xn-yn is divisible by (x-y) whenever x-y 0

Ans.P (n) : xn-yn is divisible by (x-y)

For n = 1

P (1) : x – y is divisible by (x – y)

Let P (K) be true

P (K) : xK – yK is divisible by (x – y)

( from i)

27. Prove (x2n-1) is divisible by (x-1).

Ans.P (n) : (x2n-1) is divisible by (x-1).

For n = 1

P (1) : (x2 – 1) = (x – 1) (x + 1)

which is divisible by (x – 1)

Let P (K) be true

28. Prove

Ans.P (n) :

we want to prove that p (k + 1) is true

29. Prove 1.3 + 3.5 + 5.7 + -- + (2n – 1) (2n + 1)

=

Ans.Let p (n) : 1.3 + 3.5 + -- + (2n-1) (2n+1)

=

For n = 1

P (1) = (1) (3) =

P(1) = 3 = 3 Hence p (1) is true

Let (k) be true

P(k) : 1.3 + 3.5 + -- + (2k - 1) (2K + 1) =

we want to prove that p (k+1) is true

p (k+1) : 1.3 + 3.5 + -- + (2k+1) (2k+3) =

L. H. S

Thus p (k+1) is true whenever p (k) is true.

30. Prove by PMI

3.22 + 32.23 + 33.24 + -- + 3n. 2n+1 = .

Ans.Let p (n) : 3.22 + 32.23 + 33.24 + --- + 3n. 2n+1 =

For n = 1

Thus p(k+1) is truewhenever p(k) is true.

31. Prove 1.3 + 2.32 + 3.33 + --- + n.3n =

Ans.P (n) : 1.3 + 2.32 + 3.33 + --- + n.3n =

For n = 1

P (1) : 1. 31 =

Thus p (k+1) is true whenever p(k) is true.

32. Prove

Ans. P(n) :

For n = 1

Thus p (k+1) is true whenever p (k) is true

Hence p (n) is true for all n N.

33. The sum of the cubes of three consecutive natural no. is divisible by 9.

Ans.P(n) is divisible by 9

For n = 1

P (1) : 1 + 8 + 9 = 18

which is divisible by 9

Let p (k) be true

34. Prove that 12n + 25n-1 is divisible by 13

Ans.P(n) : 12n + 25n-1 is divisible by 13

For n = 1

P(1) : 12 + (25)0 = 13

which is divisible by 13

Let p (k) be true

P(k) : 12k + 25k-1 is divisible by 13

which is divisible by 13.

35. Prove 11n+2 + 122n+1 is divisible by 133.

Ans. P(n) : 11n+2 + 122n+1 is divisible by 133.

For n = 1

P(1) : 113 + 123 = 3059

which is divisible by 133

Let p (k) be true

36. Prove 13 + 23 + 33 + --- + n3 =

Ans. P(n) : 13 + 23 + 33 + --- + n3 =

For n = 1

Thus p(k+1) is true whenever p(k) is true.

37. Prove (a)+ (a + d) + (a + 2d) + -- + [a + (n – 1)d] =

Ans. P(n) : (a)+ (a + d) + (a + 2d) + -- + [a + (n – 1)d] =

For n = 1

p(1) : a + (1-1) d = 2a + (1-1) d = a

which is true

Let p (k) be true

38. Prove that 2n > n positive integers n.

Ans. Let p (n) : 2n > n

For n = 1

P (1) : 21 >1

Which is true

Let p (k) be true

P (k) : 2k > k (1)

we want to prove that p (k+1) is true

2k> k by (1)

39. Prove

Ans. P(n) :

For n = 1

40. Prove

Ans.P(n) :

For n = 1