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#### 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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