NCERT Solutions for Class 11 Commerce Maths Chapter 4 Principle Of Mathematical Induction are provided here with simple step-by-step explanations. These solutions for Principle Of Mathematical Induction are extremely popular among Class 11 Commerce students for Maths Principle Of Mathematical Induction Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of Class 11 Commerce Maths Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s NCERT Solutions. All NCERT Solutions for class Class 11 Commerce Maths are prepared by experts and are 100% accurate.

#### 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*.

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