Class 11 NCERT Solutions- Chapter 4 Principle of Mathematical Induction – Exercise 4.1 | Set 2

Question 14:  = (n+1)

Solution:

We have,

P(n) =  = (n+1)

For n=1, we get

P(1) = (1+ ) = 2 = (1+1) = 2

So, P(1) is true

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

P(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 true for all natural numbers n.

Question 15: 1² + 3² + 5² + …… + (2n-1)² = 

Solution:

We have,

P(n) = 1² + 3² + 5² + …… + (2n-1)² = 

For n=1, we get

P(1) = 1² = 1 =  = 1

So, P(1) is true

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

P(k) = 1² + 3² + 5² + …… + (2k-1)² =  ……………..(1)

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

P(k+1) = 1² + 3² + 5² + …… + (2k-1)² + (2(k+1)-1)²

= (1² + 3² + 5² + …… + (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 true for all natural numbers n.

Question 16: 

Solution:

We have,

P(n) = 

For n=1, we get

P(1) = 

So, P(1) is true

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

P(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 true for all natural numbers n.

Question 17: 

Solution:

We have,

P(n) = 

For n=1, we get

P(1) = 

So, P(1) is true

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

P(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 true for 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 get

P(1) = 1 <  (2(1)+1)²

1 < 

So, P(1) is true

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

P(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))

<  ((4k²+4k+1) + 8k+8))

<  (4k²+12k+9)

<  (2k+3)²

P(k+1) <  (2(k+1)+1)²

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

Hence, from the principle of mathematical induction, the statement P(n) is true for 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 get

p(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=k

P(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.

Hence, from the principle of mathematical induction, the statement P(n) is true for 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 11

For n=1, we get

P (1) = 10^{2.1 – 1} + 1 = 11, which is divisible by 11

So, P(1) is true

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

10^{2k – 1} + 1 is divisible by 11

10^{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

= 10² (10^2k-1 + 1 – 1) + 1

= 10² (10^2k-1 + 1) – 10² + 1

Using Eq(1), we get

= 10². 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 true for all natural numbers i.e. n.

Question 21: x²ⁿ – y²ⁿ is divisible by x + y.

Solution:

We have,

P (n) = x²ⁿ – y²ⁿ is divisible by x + y

For n=1, we get

P (1) = x² × 1 – y² × 1 = x² – y² = (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=k

x^2k – y^2k is divisible by x + y

x^2k – y^2k = m (x + y), where m ∈ N …… (1)

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

x ^{2(k + 1)} – y ^{2(k + 1)}

= x ^2k . x² – y^2k . y²

By adding and subtracting y^2k we get

= x² (x^2k – y^2k + y^2k) – y^2k. y²

Using Eq(1), we get

= x² {m (x + y) + y^2k} – y^2k. y²

= m (x + y) x² + y^2k. x² – y^2k. y²

Taking out the common terms

= m (x + y) x² + y^2k (x² – y²)

= m (x + y) x² + y^2k (x + y) (x – y)

So we get

= (x + y) {mx² + 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: 3²ⁿ⁺² – 8n – 9 is divisible by 8.

Solution:

We have,

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

For n=1, we get

P (1) = 3^{2 × 1 + 2} – 8 × 1 – 9 = 64, which is divisible by 8

So, P(1) is true

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

3^{2k + 2} – 8k – 9 is divisible by 8

3^{2k + 2} – 8k – 9 = 8m, where m ∈ N …… (1)

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

3 ^{2(k + 1) + 2} – 8 (k + 1) – 9

= 3 ^{2k + 2} . 3² – 8k – 8 – 9

By adding and subtracting 8k and 9 we get

= 3² (3^{2k + 2} – 8k – 9 + 8k + 9) – 8k – 17

= 3² (3^{2k + 2} – 8k – 9) + 3² (8k + 9) – 8k – 17

Using 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 true for all natural numbers i.e. n.

Question 23: 41ⁿ – 14ⁿ is a multiple of 27.

Solution:

We have,

P (n) = 41ⁿ – 14ⁿ is a multiple of 27

For n=1, we get

P (1) = 41¹ – 14¹ = 27, which is a multiple by 27

So, P(1) is true

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

41^k – 14^k is a multiple of 27

41^k – 14^k = 27m, where m ∈ N …… (1)

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

41^k + 1 – 14^k + 1

= 41^k. 41 – 14^k. 14

By 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. 14

Using 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 number

So 41^{k + 1} – 14^{k + 1} is a multiple of 27

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 24:  (2n + 7) < (n + 3)².

Solution:

We have,

P(n) = (2n +7) < (n + 3)²

For n=1, we get

2.1 + 7 = 9 < (1 + 3)² = 16

So, 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 + 2

2 (k + 1) + 7 < k² + 6k + 9 + 2

2 (k + 1) + 7 < k² + 6k + 11

Here,

k² + 6k + 11 < k² + 8k + 16

2 (k + 1) + 7 < (k + 4)²

2 (k + 1) + 7 < {(k + 1) + 3}²

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.