Home » CBSE » Class 11 » NCERT Solutions for Class 11 Maths Chapter 4 – Principle of Mathematical Induction

NCERT Solutions for Class 11 Maths Chapter 4 – Principle of Mathematical Induction

NCERT Solutions for Class 11 Maths Chapter 4 – Principle of Mathematical Induction

Page No 94:

Question 1:

Prove the following by using the principle of mathematical induction for all n ∈ N:

ANSWER:

Let the given statement be P(n), i.e.,

P(n): 1 + 3 + 32 + …+ 3n–1 =

For n = 1, we have

P(1): 1 =, which is true.

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true.

Consider

1 + 3 + 32 + … + 3k–1 + 3(k+1) – 1

= (1 + 3 + 32 +… + 3k–1) + 3k

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

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Page No 94:

Question 2:

Prove the following by using the principle of mathematical induction for all n ∈ N

ANSWER:

Let the given statement be P(n), i.e.,

P(n): 

For n = 1, we have

P(1): 13 = 1 =, which is true.

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true.

Consider

13 + 23 + 33 + … + k3 + (k + 1)3

= (13 + 23 + 33 + …. + k3) + (k + 1)3 

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

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Page No 94:

Question 3:

Prove the following by using the principle of mathematical induction for all n ∈ N

ANSWER:

Let the given statement be P(n), i.e.,

P(n): 

For n = 1, we have

P(1): 1 = which is true.

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true.

Consider

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

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Page No 94:

Question 4:

Prove the following by using the principle of mathematical induction for all n ∈ N: 1.2.3 + 2.3.4 + … + n(n + 1) (n + 2) =

ANSWER:

Let the given statement be P(n), i.e.,

P(n): 1.2.3 + 2.3.4 + … + n(n + 1) (n + 2) =

For n = 1, we have

P(1): 1.2.3 = 6 =, which is true.

Let P(k) be true for some positive integer k, i.e.,

1.2.3 + 2.3.4 + … + k(k + 1) (k + 2) 

We shall now prove that P(k + 1) is true.

Consider

1.2.3 + 2.3.4 + … + k(k + 1) (k + 2) + (k + 1) (k + 2) (k + 3)

= {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.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Page No 94:

Question 5:

Prove the following by using the principle of mathematical induction for all n ∈ N

ANSWER:

Let the given statement be P(n), i.e.,

P(n) : 

For n = 1, we have

P(1): 1.3 = 3, which is true.

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true.

Consider

1.3 + 2.32 + 3.33 + … + k3k+ (k + 1) 3k+1

= (1.3 + 2.32 + 3.33 + …+ k.3k) + (k + 1) 3k+1

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

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Page No 94:

Question 6:

Prove the following by using the principle of mathematical induction for all n ∈ N

Page No 94:

Question 7:

Prove the following by using the principle of mathematical induction for all n ∈ N

ANSWER:

Let the given statement be P(n), i.e.,

P(n): 

For n = 1, we have

, which is true.

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true.

Consider

(1.3 + 3.5 + 5.7 + … + (2k – 1) (2k + 1) + {2(k + 1) – 1}{2(k + 1) + 1}

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

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Page No 94:

Question 8:

Prove the following by using the principle of mathematical induction for all n ∈ N: 1.2 + 2.22 + 3.22 + … + n.2n = (n – 1) 2n+1 + 2

ANSWER:

Let the given statement be P(n), i.e.,

P(n): 1.2 + 2.22 + 3.22 + … + n.2n = (n – 1) 2n+1 + 2

For n = 1, we have

P(1): 1.2 = 2 = (1 – 1) 21+1 + 2 = 0 + 2 = 2, which is true.

Let P(k) be true for some positive integer k, i.e.,

1.2 + 2.22 + 3.22 + … + k.2k = (k – 1) 2k + 1 + 2 … (i)

We shall now prove that P(k + 1) is true.

Consider

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

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Page No 94:

Question 9:

Prove the following by using the principle of mathematical induction for all n ∈ N

ANSWER:

Let the given statement be P(n), i.e.,

P(n): 

For n = 1, we have

P(1): , which is true.

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true.

Consider

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

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Page No 94:

Question 10:

Prove the following by using the principle of mathematical induction for all n ∈ N

ANSWER:

Let the given statement be P(n), i.e.,

P(n): 

For n = 1, we have

, which is true.

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true.

Consider

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

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Page No 94:

Question 11:

Prove the following by using the principle of mathematical induction for all n ∈ N

ANSWER:

Let the given statement be P(n), i.e.,

P(n): 

For n = 1, we have

, which is true.

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true.

Consider

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

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Page No 95:

Question 12:

Prove the following by using the principle of mathematical induction for all n ∈ N

ANSWER:

Let the given statement be P(n), i.e.,

For n = 1, we have

, which is true.

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true.

Consider

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

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Page No 95:

Question 13:

Prove the following by using the principle of mathematical induction for all n ∈ N

Page No 95:

Question 14:

Prove the following by using the principle of mathematical induction for all n ∈ N

ANSWER:

Let the given statement be P(n), i.e.,

For n = 1, we have

, which is true.

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true.

Consider

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

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Page No 95:

Question 15:

Prove the following by using the principle of mathematical induction for all n ∈ N

ANSWER:

Let the given statement be P(n), i.e.,

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true.

Consider

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

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Page No 95:

Question 16:

Prove the following by using the principle of mathematical induction for all n ∈ N

ANSWER:

Let the given statement be P(n), i.e.,

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true.

Consider

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

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Page No 95:

Question 17:

Prove the following by using the principle of mathematical induction for all n ∈ N

ANSWER:

Let the given statement be P(n), i.e.,

For n = 1, we have

, which is true.

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true.

Consider

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

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Page No 95:

Question 18:

Prove the following by using the principle of mathematical induction for all n ∈ N

ANSWER:

Let the given statement be P(n), i.e.,

It can be noted that P(n) is true for n = 1 since .

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true whenever P(k) is true.

Consider

Hence,

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

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Page No 95:

Question 19:

Prove the following by using the principle of mathematical induction for all n ∈ Nn (n + 1) (n + 5) is a multiple of 3.

ANSWER:

Let the given statement be P(n), i.e.,

P(n): n (n + 1) (n + 5), which is a multiple of 3.

It can be noted that P(n) is true for n = 1 since 1 (1 + 1) (1 + 5) = 12, which is a multiple of 3.

Let P(k) be true for some positive integer k, i.e.,

k (k + 1) (k + 5) is a multiple of 3.

k (k + 1) (k + 5) = 3m, where m ∈ N … (1)

We shall now prove that P(k + 1) is true whenever P(k) is true.

Consider

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

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Page No 95:

Question 20:

Prove the following by using the principle of mathematical induction for all n ∈ N: 102n – 1 + 1 is divisible by 11.

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ANSWER:

Let the given statement be P(n), i.e.,

P(n): 102n – 1 + 1 is divisible by 11.

It can be observed that P(n) is true for n = 1 since P(1) = 102.1 – 1 + 1 = 11, which is divisible by 11.

Let P(k) be true for some positive integer k, i.e.,

102k – 1 + 1 is divisible by 11.

∴102k – 1 + 1 = 11m, where m ∈ … (1)

We shall now prove that P(k + 1) is true whenever P(k) is true.

Consider

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

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Page No 95:

Question 21:

Prove the following by using the principle of mathematical induction for all n ∈ Nx2n – y2n is divisible by x y.

ANSWER:

Let the given statement be P(n), i.e.,

P(n): x2n – y2n is divisible by x y.

It can be observed that P(n) is true for n = 1.

This is so because x× 1 – y× 1 = x2 – y2 = (y) (x – y) is divisible by (x + y).

Let P(k) be true for some positive integer k, i.e.,

x2k – y2k is divisible by x y.

x2k – y2k = m (y), where m ∈ N … (1)

We shall now prove that P(k + 1) is true whenever P(k) is true.

Consider

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

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Page No 95:

Question 22:

Prove the following by using the principle of mathematical induction for all n ∈ N: 32n + 2 – 8n – 9 is divisible by 8.

ANSWER:

Let the given statement be P(n), i.e.,

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

It can be observed that P(n) is true for n = 1 since 3× 1 + 2 – 8 × 1 – 9 = 64, which is divisible by 8.

Let P(k) be true for some positive integer k, i.e.,

32k + 2 – 8k – 9 is divisible by 8.

∴32k + 2 – 8k – 9 = 8m; where m ∈ N … (1)

We shall now prove that P(k + 1) is true whenever P(k) is true.

Consider

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

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Page No 95:

Question 23:

Prove the following by using the principle of mathematical induction for all n ∈ N: 41n – 14n is a multiple of 27.

ANSWER:

Let the given statement be P(n), i.e.,

P(n):41n – 14nis a multiple of 27.

It can be observed that P(n) is true for n = 1 since which is a multiple of 27.

Let P(k) be true for some positive integer k, i.e.,

41k – 14kis a multiple of 27

∴41k – 14k = 27m, where m ∈ N … (1)

We shall now prove that P(k + 1) is true whenever P(k) is true.

Consider

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

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Page No 95:

Question 24:

Prove the following by using the principle of mathematical induction for all

(2+7) < (n + 3)2

ANSWER:

Let the given statement be P(n), i.e.,

P(n): (2+7) < (n + 3)2

It can be observed that P(n) is true for n = 1 since 2.1 + 7 = 9 < (1 + 3)2 = 16, which is true.

Let P(k) be true for some positive integer k, i.e.,

(2k + 7) < (k + 3)2 … (1)

We shall now prove that P(k + 1) is true whenever P(k) is true.

Consider

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

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

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