NCERT Solutions class 11 Mathematics 4. Principle Of Mathematical Induction Exercise 4.1

Chapter 4. Principle of Mathematical induction

Exercise 4.1

Prove the following by using the principle of mathematical induction for all n ∈ N:

Solution:

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

​

LHS = RHS

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. 

Solution:

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

LHS = RHS

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. 

 

Solution: Let the given statement be P(n), so

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.

Q19. n (n + 1) (n + 5) is a multiple of 3.

Solution:

Let the given statement be P(n), so

P(n) : n (n + 1) (n + 5) is a multiple of 3.

For n = 1, so we have;

n (n + 1) (n + 5) = 1 × 2 × 6 = 12 = 3 × 4

P(n) is true for n = 1

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

k(k + 1) (k + 5)

= k³ + 6k² + 5 k = 3m (say)   ……………….. (1)

Now, we shall prove that P(k + 1) is true whenever P(k) is true

Replacing k by k + 1

    k + 1 (k + 2) (k + 6)

= (k + 1) (k² + 8k + 12)

= k (k² + 8k + 12) + 1(k² + 8k + 12)

= k³ + 8k² + 12k + k² + 8k + 12

= k³ + 9k² + 20k + 12

=( k³ + 6k² + 5 k) + 3k² + 15k + 12

= 3m + 3k² + 15k + 12     from (1)

= 3(m + k² + 5k + 4)     

∴  k + 1 (k + 2) (k + 6) is 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 ∈ N.

Q20.  10^{2n - 1}  + 1 is divisible by 11.

Solution:

Let the given statement be P(n), so

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

For n = 1, so we have;

10^{2n - 1}  + 1 = 10^{2×1 - 1} + 1 = 10 + 1 = 11

P(n) is true for n = 1

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

10^{2k- 1}  + 1  = 11m say

10^{2k- 1}  = 11m - 1      ……………… (1)

We shall prove that P(k + 1) is true whenever P(k) is true

∴ replacing k by k + 1 we have

     10^{2k - 1}  + 1

=  10^{2k + 1}  + 1

= 10^2k × 10¹ + 1 

= {10^{2k - 1} × 100 + 1}  

= {(11m - 1)× 100 + 1}     from equation (1)

= 1100m - 100+ 1   

= 1100m - 99

= 11(100m - 9)     

∴ 10^{2n - 1}  + 1 is divisible by 11

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 ∈ N.

Q21.  x²ⁿ – y²ⁿ is divisible by x + y

Solution: Let the given statement be P(n), so

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

Putting n = 1 we have,

x²ⁿ – y²ⁿ = x² - y² = (x + y) (x - y)

P(n) is true for n = 1

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

x^2k – y^2k is divisible by (x + y)

So, x^2k – y^2k = m( x + y)

Or  x^2k = m( x + y) + y^2k     …………. (1)

We shall prove that P(k + 1) is true whenever P(k) is true

∴ replacing k by k + 1 we have

    x^{2k + 2} – y^{2k + 2}

= x^2k . x²  – y^2k .y²

Putting the value of x^2k from (1)

= {m( x + y) + y^2k} x²  – y^2k .y²

= m( x + y) x² + y^2k. x²  – y^2k .y²

= m( x + y) x² + y^2k (x²  – y²)

= m( x + y) x² + y^2k (x + y) ( x - y)

= ( x + y) [mx² + y^2k ( x - y)]

∴ x²ⁿ – y²ⁿ is divisible by x + y

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 ∈ N.

Q22.  3²ⁿ⁺² – 8n – 9 is divisible by 8

Solution: Let the given statement be P(n), so

P(n) : 3²ⁿ⁺² – 8n – 9 is divisible by 8
Putting n =1

P(1) : 3^2×1+2 – 8 × 1 – 9 = 81 - 17 = 64 = 8 × 8

Which is divisible by 8

∴ P(1) is true

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

 3^{2k + 2} – 8k – 9

 3^{2k + 2} – 8k – 9 is divisible by 8
 3^{2k + 2} – 8k – 9 = 8m

Or 3^{2k + 2} = 8m + 8k + 9     ……………. (1)

We shall prove that P(k + 1) is true whenever P(k) is true

∴ replacing k by k + 1 we have

    3^{2k + 4} – 8k – 8  – 9

= 3^{2k + 4} – 8k – 17

= 3^{2k + 2} × 3² – 8k – 17

= (8m + 8k + 9)× 9 – 8k – 17

= 72m + 72k + 81 – 8k – 17

= 72m + 64k + 64

= 8(9m + 8k + 8)

∴  3²ⁿ⁺² – 8n – 9 is divisible by 8

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 ∈ N.

Q23. 41ⁿ – 14ⁿ is a multiple of 27.

Solution: Let the given statement be P(n), so

P(n) : 41ⁿ – 14ⁿ is a multiple of 27
Putting n = 1

P(1): 41ⁿ – 14ⁿ = 41 – 14 = 27

∴ P(1) is true

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

41^k – 14^k = 27

41^k = 27 + 14^k   ………… (1)

We shall prove that P(k + 1) is true whenever P(k) is true

∴ replacing k by k + 1 we have

     41^{k + 1} – 14^{k + 1}

=  41^k . 41 – 14^k . 14

=  (27 + 14^k) 41 – 14^k . 14

=  27 . 41 + 14^k .41 – 14^k . 14

=  27 . 41 + 14^k (41 – 14 )

=  27 . 41 + 14^k . 27

=  27 ( 41 + 14^k )

∴ 41ⁿ – 14ⁿ is a multiple of 27
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 ∈ N.

Q24. (2n + 7) < (n + 3)²

Solution: Let the statement be p(n) so,

p(n) : (2n + 7) < (n + 3)²

=> p(1) :  (2 × 1 + 7) < (1 + 3)²

=> 9 < 4²

=> 9 < 16 

Therefore, p(1) is true so Assume that p(k) is also true for some integer k. 

(2k + 7) < (k + 3)²  ......... (i) 

Now we shall prove for p(k + 1) 

2(k +1) + 7 < (k + 1 + 3)²

2k + 2 + 7 < (k + 4)²  ........ (ii) 

We have from (i) 

(2k + 7) < (k + 3)² 

Adding 2 both sides

=> 2k + 7 + 2 < (k + 3)² + 2 

=> 2k + 7 + 2 < k² + 6k + 9 + 2 

=> 2k + 7 + 2 < k²​ + 6k + 9 + 2 

=> 2k + 7 + 2 < k²​ + 6k + 11

Now, k²​ + 6k + 11 < (k + 4)² from (ii) 

=> 2k + 7 + 2 < k² + 6k + 11 < k² + 8k + 16 

=> 2k + 2 + 7 < k² + 8k + 16

=> 2(k + 1) + 7 < (k + 4)²

=> 2(k + 1) + 7 < (k + 1 + 3)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 ∈ N.