Exercise 4.1 Class 11 maths 4. Principle Of Mathematical Induction - ncert solutions - Toppers Study

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.