NCERT Solutions for Class 11 Maths Chapter 4

Principle of Mathematical Induction Class 11

Chapter 4 Principle of Mathematical Induction Exercise 4.1 Solutions

 Exercise 4.1 : Solutions of Questions on Page Number : 94                                                                                                                 

Q1 :

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

Answer :

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

P(n): 1 + 3 + 32 + …+ 3n–1 =

For n = 1, we have

P(1): 1 = , which is true.

We shall now prove that P(k + 1) is true. Consider

1 + 3 + 32 + … + 3k–1 + 3(k+1) – 1

= (1 + 3 + 32 +… + 3k–1) + 3k

www.ncrtsolutions.in

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

Q2 :

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

Answer :

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

P(n):

For n = 1, we have

P(1): 13 = 1 = , which is true.

We shall now prove that P(k + 1) is true. Consider

13 + 23 + 33 + … + k3 + (k + 1)3

= (13 + 23 + 33 + …. + k3) + (k + 1)3

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

Q3 :

Prove the following by using the principle of mathematical induction for

all n ∈ N:   Answer :

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

P(n):

For n = 1, we have

P(1): 1 =   which is true.

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true. Consider

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

Q4 :

Prove the following by using the principle of mathematical induction for all n ∈ N: 1.2.3 + 2.3.4 + … + n(n + 1) (n + 2) =

Answer :

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

P(n): 1.2.3 + 2.3.4 + … + n(n + 1) (n + 2) =

For n = 1, we have

P(1): 1.2.3 = 6 = , which is true.

Let P(k) be true for some positive integer k, i.e.,

1.2.3 + 2.3.4 + … + k(k + 1) (k + 2)

We shall now prove that P(k + 1) is true. Consider

1.2.3 + 2.3.4 + … + k(k + 1) (k + 2) + (k + 1) (k + 2) (k + 3)

= {1.2.3 + 2.3.4 + … + k(k + 1) (k + 2)} + (k + 1) (k + 2) (k + 3)

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

true.

Q5 :

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

Answer :

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

P(n) :

For n = 1, we have

P(1): 1.3 = 3 , which is true.

We shall now prove that P(k + 1) is true. Consider

1.3 + 2.32 + 3.33 + … + k3k+ (k + 1) 3k+1

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

Q6 :

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

Answer :

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

P(n):

For n = 1, we have

P(1): , which is true.

We shall now prove that P(k + 1) is true. Consider

1.2 + 2.3 + 3.4 + … + k.(k + 1) + (k + 1).(k + 2)

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

Q7 :

Prove the following by using the principle of mathematical induction for

all n ∈ N:   Answer :

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

P(n):

For n = 1, we have

, which is true.

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

Consider

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

Q8 :

Prove the following by using the principle of mathematical induction for all n ∈ N: 1.2 + 2.22 + 3.22 + … + n.2n= (n - 1) 2n+1 + 2

Answer :

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

P(n): 1.2 + 2.22 + 3.22 + … + n.2n = (n – 1) 2n+1 + 2

For n = 1, we have

P(1): 1.2 = 2 = (1 – 1) 21+1 + 2 = 0 + 2 = 2, which is true. Let P(k) be true for some positive integer k, i.e.,

1.2 + 2.22 + 3.22 + … + k.2k = (k – 1) 2k + 1 + 2 … (i)

We shall now prove that P(k + 1) is true. Consider

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

Q9 :

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

Answer :

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

P(n):

For n = 1, we have

P(1): , which is true.

We shall now prove that P(k + 1) is true. Consider

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

Q10 :

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

Answer :

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

P(n):

For n = 1, we have

We shall now prove that P(k + 1) is true. Consider

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

Q11 :

Prove the following by using the principle of mathematical induction for

all n ∈ N:   Answer :

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

P(n):

For n = 1, we have

, which is true.

We shall now prove that P(k + 1) is true. Consider

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

Q12 :

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

For n = 1, we have

, which is true.

We shall now prove that P(k + 1) is true. Consider

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

Q13 :

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

For n = 1, we have

We shall now prove that P(k + 1) is true. Consider

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

Q14 :

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

Answer :

For n = 1, we have

We shall now prove that P(k + 1) is true. Consider

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

Q15 :

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

Answer :

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true. Consider

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

Q16 :

Prove the following by using the principle of mathematical induction for

all n ∈ N:   Answer :

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

We shall now prove that P(k + 1) is true. Consider

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

Q17 :

Prove the following by using the principle of mathematical induction for

all n ∈ N:   Answer :

For n = 1, we have

, which is true. Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true. Consider

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

Q18 :

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

Answer :

It can be noted that P(n) is true for n = 1 since . Let P(k) be true for some positive integer k, i.e.,

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

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

Q19 :

Prove the following by using the principle of mathematical induction for all n ∈ N: n (n + 1) (n + 5) is a multiple of 3.

Answer :

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

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

It can be noted that P(n) is true for n = 1 since 1 (1 + 1) (1 + 5) = 12, which is a multiple of 3. Let P(k) be true for some positive integer k, i.e.,

k (k + 1) (k + 5) is a multiple of 3.

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

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

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

Q20 :

Prove the following by using the principle of mathematical induction for all n ∈ N: 102n - 1 + 1 is divisible by 11.

Answer :

Let the given statement be P(n), i.e., P(n): 102n – 1 + 1 is divisible by 11.

It can be observed that P(n) is true for n = 1 since P(1) = 102.1 – 1  + 1 = 11, which is divisible by 11. Let P(k) be true for some positive integer k, i.e.,

102k– 1 + 1 is divisible by 11.

∴102k– 1 + 1 = 11m, where m ∈ N … (1)

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

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

Q21 :

Prove the following by using the principle of mathematical induction for all n ∈ N: x2n - y2n is divisible by x+ y.

Answer :

Let the given statement be P(n), i.e., P(n): x2n – y2n is divisible by x + y.

It can be observed that P(n) is true for n = 1.

This is so because x2 ×1 – y2 ×1 = x2 – y2 = (x + y) (x – y) is divisible by (x + y). Let P(k) be true for some positive integer k, i.e.,

x2k – y2k is divisible by x + y.

∴x2k – y2k = m (x + y), where m ∈ N … (1)

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

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

Q22 :

Prove the following by using the principle of mathematical induction for all n ∈ N: 32n + 2 - 8n - 9 is divisible by 8.

Answer :

Let the given statement be P(n), i.e., P(n): 32n + 2 – 8n – 9 is divisible by 8.

It can be observed that P(n) is true for n = 1 since 32  ×1 + 2 – 8 × 1 – 9 = 64, which is divisible by 8. Let P(k) be true for some positive integer k, i.e.,

32k+ 2 – 8k – 9 is divisible by 8.

∴32k+ 2 – 8k – 9 = 8m; where m ∈ N … (1)

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

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

Q23 :

Prove the following by using the principle of mathematical induction for all n ∈ N: 41n - 14n is a multiple of 27.

Answer :

Let the given statement be P(n), i.e., P(n):41n – 14nis a multiple of 27.

It can be observed that P(n) is true for n = 1 since , which is a multiple of 27.

Let P(k) be true for some positive integer k, i.e., 41k – 14kis a multiple of 27

∴41k – 14k = 27m, where m ∈ N … (1)

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

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

Q24 :

Prove the following by using the principle of mathematical induction for all  (2n +7) < (n + 3)2

Answer :

Let the given statement be P(n), i.e., P(n): (2n +7) < (n + 3)2

It can be observed that P(n) is true for n = 1 since 2.1 + 7 = 9 < (1 + 3)2 = 16, which is true. Let P(k) be true for some positive integer k, i.e.,

(2k + 7) < (k + 3)2 … (1)

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

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

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.