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

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

**4 Marks Questions**

**1. For every integer n, prove that **

**Ans.** P(n) : 7^{n}-3^{n} is divisible by 4

For n=1

P(1) : 7^{1 }-3^{1} =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+2.2 ^{2}+3.2^{3}+…+n.2^{n }= (n-1)2 ^{n + 1}+2**

**Ans. **

**c**

**9. Prove that 2.7 ^{n} + 3.5^{n} – 5 is divisible by 24 **

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

for n = 1

P (1) : 2.7^{1} + 3.5^{1} – 5 = 24 is divisible by 24

Hence result is true for n = 1

Let P (K) be true

P (K) : 2.7^{K} + 3.5^{K} – 5

(i)

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

2.7^{K+1} + 3.5^{K+1} – 5 = 2.7^{K} .7^{1} + 3.5^{K} . 5^{1} -5

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

**10. Prove that 41 ^{n} – 14^{n} is a multiple of 27 **

**Ans. **P (n) : 41^{n} – 14^{n} is a multiple of 27

for n = 1

P (1) : 41^{1} – 14 = 27, which is a multiple of 27

Let P (K) be True

P (K) : 41^{K} – 14^{K}

(i)

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

41^{K+1}– 14^{k+1} = 41^{K}.41-14^{K}. 14

is a multiple of 27

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

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

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

For n = 1

p (1) : 10^{1} + 3.4^{1+2} + 5=207, divisible by 9

Hence result is true for n = 1

Let p (K) be true

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

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

10^{(K+1)} +3.4^{K+1+2}+5=10^{K+1}+3.4^{K+3}+5

which is divisible by 9.

**12. Prove that 1 ^{2} + 2^{2} + 3^{2} + ---+ n^{2} = **

**Ans. **Let P(n) : 1^{2} + 2^{2} + 3^{2} + -- +n^{2} =

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+3 ^{2} + -- + 3^{n-1 }= **

**Ans**. Let

P (n) : 1 + 3 + 3^{2} + -- + 3^{n-1} = for n = 1

P (1) 3^{1-1 }=

which is true

Let P (K) be true

P (K) : 1 + 3 + 3^{2} + ---- + 3^{K-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 1 ^{2} + 2^{2} + 3^{2} + -- + n^{2} > **

**Ans. **Let P (n) : 1^{2} + 2^{2} + 3^{2} + --- + n^{2} >

for n = 1

12 > which is true

Let P (K) be true

P (K) : 1^{2} + 2^{2} + 3^{2} + -- + K^{2} >

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

P (K+1) : 1^{2} + 2^{2} + --- + (K+1)^{2}

Hence by PMI P (n) is true

**15. Prove by PMI (ab) ^{n} = a^{n} b^{n}**

**Ans. **Let P (n) : (ab)^{n} = a^{n} b^{n}

for n = 1

ab = ab which is true

Let P (K) be true

(ab)^{K} = a^{K} b^{K} (1)

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

(ab)^{K+1} = a^{K+1}. b^{K+1}

L.H.S = (ab)^{K+1}

**16. Prove by PMI a + ar + ar ^{2} + ---- + ar^{n-1} = **

**Ans. **Let P (n) : a + ar + ar^{2} + -- + ar^{n-1} =

for n = 1

P (1) = a = a which is true

Let P (K) be true

P (K) : a + ar + ar^{2} + -- + ar^{K-1} =

we want to prove that

P (K+1) : a + ar + ar^{2} + -- + ar^{K} =

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 x ^{2n} – y^{2n} is divisible by x + y.**

**Ans. **P (n) : x^{2n} – y^{2n} is divisible by x + y

for n = 1

p (1) : x^{2} – y^{2} = (x – y) (x + y), which is divisible by x + y

Hence result is true for n = 1

Let P (K) be true

p (K) : x^{2K} – y^{2K} 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 2 ^{3n} – 1 is divisible by 7 . **

**Ans.**P (n) : 2^{3n} – 1 is divisible by 7

for n = 1

P (1) : 2^{3} – 1 = 7 which is divisible by 7

Let P (K) be true

P (K) : 2^{3K} – 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 n ^{2}. **

**Ans.**Let P (n) : 1 + 3 + 5 + --- + (2n-1) = n^{2}

For n = 1

P (1) = 1 = 1 which is true

Let P (K) be true

P (K) : 1 + 3 + 5 + --- + (2K-1) = K^{2} (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) : 1^{3} = 1^{3} 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 3 ^{2n+2} – 8n – 9 is divisible by 8 **

**Ans.**P(n) : 3^{2n+2} – 8n – 9 is divisible by 8

For n = 1

P (1) : 3^{2+2} – 8 – 9 = 64

which is divisible by 8

Hence result is true for n = 1

Let P (K) be true

P (K) : 3^{2K+2} – 8K – 9 is divisible by 8

( from i)

**26. Prove by PMI.**

**x ^{n}-y^{n} is divisible by (x-y) whenever x-y 0**

**Ans.**P (n) : x^{n}-y^{n} is divisible by (x-y)

For n = 1

P (1) : x – y is divisible by (x – y)

Let P (K) be true

P (K) : x^{K} – y^{K} is divisible by (x – y)

( from i)

**27. Prove (x ^{2n}-1) is divisible by (x-1).**

**Ans.**P (n) : (x^{2n}-1) is divisible by (x-1).

For n = 1

P (1) : (x^{2} – 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 **

**3.2 ^{2} + 3^{2}.2^{3} + 3^{3}.2^{4} + -- + 3^{n}. 2^{n+1} = .**

**Ans.**Let p (n) : 3.2^{2} + 3^{2}.2^{3} + 3^{3}.2^{4} + --- + 3^{n}. 2^{n+1} =

For n = 1

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

**31. Prove 1.3 + 2.3 ^{2} + 3.3^{3} + --- + n.3^{n} = **

**Ans.**P (n) : 1.3 + 2.3^{2} + 3.3^{3} + --- + n.3^{n} =

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 12 ^{n} + 25^{n-1} is divisible by 13 **

**Ans.**P(n) : 12^{n} + 25^{n-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) : 12^{k} + 25^{k-1} is divisible by 13

which is divisible by 13.

**35. Prove 11 ^{n+2} + 12^{2n+1} is divisible by 133. **

**Ans. **P(n) : 11^{n+2} + 12^{2n+1} is divisible by 133.

For n = 1

P(1) : 11^{3} + 12^{3} = 3059

which is divisible by 133

Let p (k) be true

**36. Prove 1 ^{3} + 2^{3} + 3^{3} + --- + n^{3} = **

**Ans. **P(n) : 1^{3} + 2^{3} + 3^{3} + --- + n^{3} =

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 2 ^{n} > n positive integers n. **

**Ans. **Let p (n) : 2^{n} > n

For n = 1

P (1) : 2^{1} >1

Which is true

Let p (k) be true

P (k) : 2^{k} > k (1)

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

2^{k}> k by (1)

**39. Prove **

**Ans. **P(n) :

For n = 1

**40. Prove **

**Ans.**P(n) :

For n = 1