Study Materials

NCERT Solutions for Class 11th Mathematics

 

Chapter 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:

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.  102n - 1  + 1 is divisible by 11.

Solution:

Let the given statement be P(n), so

P(n) : 102n - 1  + 1 is divisible by 11.

For n = 1, so we have;

102n - 1  + 1 = 102×1 - 1 + 1 = 10 + 1 = 11

P(n) is true for n = 1

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

102k- 1  + 1  = 11m say

102k- 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

     102k - 1  + 1

=  102k + 1  + 1

= 102k × 101 + 1 

= {102k - 1 × 100 + 1}  

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

= 1100m - 100+ 1   

= 1100m - 99

= 11(100m - 9)     

102n - 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

x2k – y2k is divisible by (x + y)

So, x2k – y2k = m( x + y)

Or  x2k = m( x + y) + y2k     …………. (1)

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

replacing k by k + 1 we have

    x2k + 2 – y2k + 2

= x2k . x2  – y2k .y2

Putting the value of x2k from (1)

= {m( x + y) + y2k} x2  – y2k .y2

= m( x + y) x2 + y2k. x2  – y2k .y2

= m( x + y) x2 + y2k (x2  – y2)

= m( x + y) x2 + y2k (x + y) ( x - y)

= ( x + y) [mx2 + y2k ( 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) : 32×1+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

 32k + 2 – 8k – 9

 32k + 2 – 8k – 9 is divisible by 8
 32k + 2 – 8k – 9 = 8m

Or 32k + 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

    32k + 4 – 8k 8  – 9

= 32k + 4 – 8k 17

= 32k + 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

41k – 14k = 27

41k = 27 + 14k   ………… (1)

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

replacing k by k + 1 we have

     41k + 1 – 14k + 1

=  41k . 41 – 14k . 14

=  (27 + 14k) 41 – 14k . 14

=  27 . 41 + 14k .41 – 14k . 14

=  27 . 41 + 14k (41 – 14 )

=  27 . 41 + 14k . 27

=  27 ( 41 + 14k )

 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 < k+ 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(+ 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.

 

 

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