## 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 + 3^{2} + …+ 3^{n}^{–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 + 3^{2} + … + 3^{k}^{–1} + 3^{(}^{k}^{+1) – 1}

= (1 + 3 + 3^{2} +… + 3^{k}^{–1}) + 3^{k}

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): 1^{3} = 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} + 2^{3} + 3^{3} + … + *k*^{3} + (*k* + 1)^{3}

= (1^{3} + 2^{3} + 3^{3} + …. + *k*^{3}) + (*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.

*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 + 2.3^{2} + 3.3^{3} + … + *k*3^{k}+ (*k* + 1) 3^{k}^{+1}

= (1.3 + 2.3^{2} + 3.3^{3} + …+ *k.*3^{k}) + (*k* + 1) 3^{k}^{+1}

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

*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*:

**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

1.2 + 2.3 + 3.4 + … + *k*.(*k *+ 1) + (*k* + 1).(*k* + 2)

= [1.2 + 2.3 + 3.4 + … + *k*.(*k* + 1)] + (*k* + 1).(*k* + 2)

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

*n*) is true for all natural numbers i.e., *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 + … + (2*k* – 1) (2*k* + 1) + {2(*k* + 1) – 1}{2(*k* + 1) + 1}

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

*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 + 2.2^{2} + 3.2^{2} + … + *n*.2^{n} = (*n* – 1) 2^{n}^{+1} + 2

**ANSWER:**

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

P(*n*): 1.2 + 2.2^{2} + 3.2^{2} + … + *n*.2^{n} = (*n* – 1) 2^{n}^{+1} + 2

For *n* = 1, we have

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

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

1.2 + 2.2^{2} + 3.2^{2} + … + *k.*2^{k} = (*k* – 1) 2^{k}^{ + 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.

*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.

*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.

*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.

*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.

*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*:

**ANSWER:**

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

For *n* = 1, we have

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.

*n*) is true for all natural numbers i.e., *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.

*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.

*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.

*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.

*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.

*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* ∈ *N*: *n* (*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) = 3*m*, 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.

*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*: 10^{2}^{n}^{ – 1 }+ 1 is divisible by 11.

**ANSWER:**

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

P(*n*): 10^{2}^{n}^{ – 1 }+ 1 is divisible by 11.

It can be observed that P(*n*) is true for *n* = 1 since P(1) = 10^{2.1 – 1 }+ 1 = 11, which is divisible by 11.

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

10^{2}^{k}^{ – 1 }+ 1 is divisible by 11.

∴10^{2}^{k}^{ – 1 }+ 1 = 11*m*, 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.

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

**ANSWER:**

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

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

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

This is so because *x*^{2 }^{×}^{ 1} – *y*^{2 }^{×}^{ 1} = *x*^{2} – *y*^{2} = (*x *+ *y*) (*x* – *y*) is divisible by (*x* + *y*).

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

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

∴*x*^{2}^{k} – *y*^{2}^{k} = *m* (*x *+ *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.

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

**ANSWER:**

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

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

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

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

3^{2}^{k}^{ + 2} – 8*k* – 9 is divisible by 8.

∴3^{2}^{k}^{ + 2} – 8*k* – 9 = 8*m*; 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.

*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*: 41^{n} – 14^{n} is a multiple of 27.

**ANSWER:**

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

P(*n*):41^{n} – 14^{n}is 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.,

41^{k} – 14^{k}is a multiple of 27

∴41^{k} – 14^{k} = 27*m*, 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.

*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*n *+7) < (*n* + 3)^{2}

**ANSWER:**

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

P(*n*): (2*n *+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.,

(2*k* + 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.

*n*) is true for all natural numbers i.e., *n*.