### Question 14: = (n+1)

**Solution:**

We have,

P(n) = = (n+1)

For

n=1, we getP(1) = (1+ ) = 2 = (1+1) = 2

So, P(1) is true

Assume that P(k) is true for some positive integer

n=kP(k) = = (k+1) ……………..(1)

Let’s prove that P(k + 1) is also true. Now, we have

P(k+1)== ()

From Eq(1), we get

= (k+1) (1+ )

= (k+1)

= {(k+1)+1}

Hence,

P(k+1) = {(k+1)+1}

Thus P(k + 1) is true, whenever P (k) is true.Hence, from the principle of mathematical induction, the statement P(n) is

truefor all natural numbers n.

### Question 15: 1^{2} + 3^{2} + 5^{2} + …… + (2n-1)^{2} =

**Solution:**

We have,

P(n) = 1

^{2}+ 3^{2}+ 5^{2}+ …… + (2n-1)^{2}=For

n=1, we getP(1) = 1

^{2}= 1 = = 1So, P(1) is true

Assume that P(k) is true for some positive integer

n=kP(k) = 1

^{2}+ 3^{2}+ 5^{2}+ …… + (2k-1)^{2}= ……………..(1)Let’s prove that P(k + 1) is also true. Now, we have

P(k+1)= 1^{2}+ 3^{2}+ 5^{2}+ …… + (2k-1)^{2}+ (2(k+1)-1)^{2}= (1

^{2}+ 3^{2}+ 5^{2}+ …… + (2k-1)^{2}) + (2k+1)^{2}From Eq(1), we get

= + (2k+1)

^{2}= (2k+1)

= (2k+1)

= (2k+1)

= (2k+1)

=

=

Hence,

P(k+1) =

Thus P(k + 1) is true, whenever P (k) is true.Hence, from the principle of mathematical induction, the statement P(n) is

truefor all natural numbers n.

### Question 16:

**Solution:**

We have,

P(n) =

For

n=1, we getP(1) =

So, P(1) is true

Assume that P(k) is true for some positive integer

n=kP(k) = ……………..(1)

Let’s prove that P(k + 1) is also true. Now, we have

P(k+1)== ()

From Eq(1), we get

=

=

=

=

=

=

Hence,

P(k+1) =

Thus P(k + 1) is true, whenever P (k) is true.Hence, from the principle of mathematical induction, the statement P(n) is

truefor all natural numbers n.

### Question 17:

**Solution:**

We have,

P(n) =

For

n=1, we getP(1) =

So, P(1) is true

Assume that P(k) is true for some positive integer

n=kP(k) = ……………..(1)

Let’s prove that P(k + 1) is also true. Now, we have

P(k+1)== ()

From Eq(1), we get

=

=

=

=

=

=

=

Hence,

P(k+1) =

Thus P(k + 1) is true, whenever P (k) is true.truefor all natural numbers n.

### Question 18: 1 + 2 + 3 + ….. + n < (2n+1)^{2}

**Solution:**

We have,

P(n) = 1 + 2 + 3 + ….. + n < (2n+1)

^{2}For

n=1, we getP(1) = 1 < (2(1)+1)

^{2}1 <

So, P(1) is true

Assume that P(k) is true for some positive integer

n=kP(k) = 1 + 2 + 3 + ….. + k < (2k+1)

^{2}……………..(1)Let’s prove that P(k + 1) is also true. Now, we have

P(k+1)= 1 + 2 + 3 + ….. + k + (k+1)= (1 + 2 + 3 + ….. + k) + k+1 < (2k+1)

^{2}+ (k+1)< ((2k+1)

^{2}+ 8(k+1))< ((4k

^{2}+4k+1) + 8k+8))< (4k

^{2}+12k+9)< (2k+3)

^{2}P(k+1) < (2(k+1)+1)

^{2}

Thus P(k + 1) is true, whenever P (k) is true.truefor all natural numbers n.

### Question 19: n (n + 1) (n + 5) is a multiple of 3.

**Solution:**

We have,

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

For

n=1, we getp(1) = 1 (1 + 1) (1 + 5) = 12, which is a multiple of 3

So, P(1) is true

Assume that P(k) is true for some positive integer

n=kP(k) = k (k + 1) (k + 5) is a multiple of 3

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

Let’s prove that P(k + 1) is also true. Now, we have

P(k+1)= (k + 1) {(k + 1) + 1} {(k + 1) + 5}= (k + 1) (k + 2) {(k + 5) + 1}

By multiplying the terms

= (k + 1) (k + 2) (k + 5) + (k + 1) (k + 2)

= {k (k + 1) (k + 5) + 2 (k + 1) (k + 5)} + (k + 1) (k + 2)

Using Eq(1), we get

= 3m + (k + 1) {2 (k + 5) + (k + 2)}

= 3m + (k + 1) {2k + 10 + k + 2}

= 3m + (k + 1) (3k + 12)

= 3m + 3 (k + 1) (k + 4)

= 3 {m + (k + 1) (k + 4)}

= 3 × q (where q = {m + (k + 1) (k + 4)} is some natural number)

(k + 1) {(k + 1) + 1} {(k + 1) + 5} is a multiple of 3

Thus P(k + 1) is true, whenever P (k) is true.truefor all natural numbers n.

### Question 20: 10^{2n – 1} + 1 is divisible by 11.

**Solution:**

We have,

P (n): 10

^{2n – 1}+ 1 is divisible by 11For

n=1, we getP (1) = 10

^{2.1 – 1}+ 1 = 11, which is divisible by 11So, P(1) is true

Assume that P(k) is true for some positive integer

n=k10

^{2k – 1}+ 1 is divisible by 1110

^{2k – 1}+ 1 = 11m, where m ∈ N …………… (1)Let’s prove that P(k + 1) is also true. Now, we have

P(k + 1)= 10^{2(k + 1) – 1}+ 1= 10

^{2k + 2 – 1}+ 1= 10

^{2k + 1}+ 1= 10

^{2}(10^{2k-1}+ 1 – 1) + 1= 10

^{2}(10^{2k-1}+ 1) – 10^{2}+ 1Using Eq(1), we get

= 10

^{2}. 11m – 100 + 1= 100 × 11m – 99

Taking out the common terms

= 11 (100m – 9)

= 11r, where r = (100m – 9) is some natural number

10 2

^{(k + 1) – 1}+ 1 is divisible by 11

P (k + 1) is true whenever P (k) is true.Therefore, by the principle of mathematical induction, statement P (n) is

truefor all natural numbers i.e. n.

### Question 21: x^{2n} – y^{2n} is divisible by x + y.

**Solution:**

We have,

P (n) = x

^{2n}– y^{2n}is divisible by x + yFor

n=1, we getP (1) = x

^{2}× 1 – y^{2}× 1 = x^{2}– y^{2}= (x + y) (x – y), which is divisible by (x + y)So, P(1) is true

Assume that P(k) is true for some positive integer

n=kx

^{2k}– y^{2k}is divisible by x + yx

^{2k}– y^{2k}= m (x + y), where m ∈ N …… (1)Let’s prove that

P(k + 1)is also true. Now, we havex

^{2(k + 1)}– y^{2(k + 1)}= x

^{2k}. x^{2}– y^{2k}. y^{2}By adding and subtracting y

^{2k}we get= x

^{2}(x^{2k}– y^{2k}+ y^{2k}) – y^{2k}. y^{2}Using Eq(1), we get

= x

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

^{2}+ y^{2k}. x^{2}– y^{2k}. y^{2}Taking out the common terms

= m (x + y) x

^{2}+ y^{2k}(x^{2}– y^{2})= m (x + y) x

^{2}+ y^{2k}(x + y) (x – y)So we get

= (x + y) {mx

^{2}+ y^{2k}(x – y)}, which is a factor of (x + y)

P (k + 1) is true whenever P (k) is true.Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

### Question 22: 3^{2n+2} – 8n – 9 is divisible by 8.

**Solution:**

We have,

P (n) = 3

^{2n + 2}– 8n – 9 is divisible by 8For

n=1, we getP (1) = 3

^{2 × 1 + 2}– 8 × 1 – 9 = 64, which is divisible by 8So, P(1) is true

Assume that P(k) is true for some positive integer

n=k3

^{2k + 2}– 8k – 9 is divisible by 83

^{2k + 2}– 8k – 9 = 8m, where m ∈ N …… (1)Let’s prove that

P(k + 1)is also true. Now, we have3

^{2(k + 1) + 2}– 8 (k + 1) – 9= 3

^{2k + 2}. 3^{2}– 8k – 8 – 9By adding and subtracting 8k and 9 we get

= 3

^{2}(3^{2k + 2}– 8k – 9 + 8k + 9) – 8k – 17= 3

^{2}(3^{2k + 2}– 8k – 9) + 3^{2}(8k + 9) – 8k – 17Using Eq(1), we get

= 9. 8m + 9 (8k + 9) – 8k – 17

= 9. 8m + 72k + 81 – 8k – 17

= 9. 8m + 64k + 64

By taking out the common terms

= 8 (9m + 8k + 8)

= 8r, where r = (9m + 8k + 8) is a natural number

So 3

^{2(k + 1) + 2}– 8 (k + 1) – 9 is divisible by 8

P (k + 1) is true whenever P (k) is true.Therefore, by the principle of mathematical induction, statement P (n) is

truefor all natural numbers i.e. n.

### Question 23: 41^{n} – 14^{n} is a multiple of 27.

**Solution:**

We have,

P (n) = 41

^{n}– 14^{n}is a multiple of 27For

n=1, we getP (1) = 41

^{1}– 14^{1}= 27, which is a multiple by 27So, P(1) is true

Assume that P(k) is true for some positive integer

n=k41

^{k}– 14^{k }is a multiple of 2741

^{k}– 14^{k}= 27m, where m ∈ N …… (1)Let’s prove that

P(k + 1)is also true. Now, we have41

^{k}+ 1 – 14^{k}+ 1= 41

^{k}. 41 – 14^{k}. 14By adding and subtracting 14

^{k}we get= 41 (41

^{k}– 14^{k}+ 14^{k}) – 14^{k}. 14= 41 (41

^{k}– 14^{k}) + 41. 14^{k}– 14^{k}. 14Using Eq(1), we get

= 41. 27m + 14

^{k}( 41 – 14)= 41. 27m + 27. 14

^{k}By taking out the common terms

= 27 (41m – 14k)

= 27r, where r = (41m – 14

^{k}) is a natural numberSo 41

^{k + 1}– 14^{k + 1}is a multiple of 27

P (k + 1) is true whenever P (k) is true.truefor all natural numbers i.e. n.

### Question 24: (2n + 7) < (n + 3)^{2}.

**Solution:**

We have,

P(n) = (2n +7) < (n + 3)

^{2}For

n=1, we get2.1 + 7 = 9 < (1 + 3)

^{2}= 16So, P(1) is true

Assume that P(k) is true for some positive integer

n=k(2k + 7) < (k + 3)

^{2}… (1)Let’s prove that

P(k + 1)is also true. Now, we have{2 (k + 1) + 7} = (2k + 7) + 2

= {2 (k + 1) + 7}

Using Eq(1), we get

(2k + 7) + 2 < (k + 3)

^{2}+ 22 (k + 1) + 7 < k

^{2}+ 6k + 9 + 22 (k + 1) + 7 < k

^{2}+ 6k + 11Here,

k

^{2}+ 6k + 11 < k^{2}+ 8k + 162 (k + 1) + 7 < (k + 4)

^{2}2 (k + 1) + 7 < {(k + 1) + 3}

^{2}

P (k + 1) is true whenever P (k) is true.truefor all natural numbers i.e. n.