NCERT Solutions Class 11 Mathematics Chapter 4 Principle of Mathematical Induction

Exercise 4.1

1. Prove the following by using the principle of mathematical induction for all n ∈ N:
1 + 3 + 1 + 3 + 3² + ….. + 3^n-1 = (3ⁿ – 1)/2

Answer :

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

P(n): 1 + 3 + 3² + …. + 3^n-1 = (3ⁿ – 1)/2 

For n = 1 we have 

P(1): 1 = (3¹ – 1)/2 = (3-1)/2 = 2/2 = 1, which is true. 

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

1 + 3 + 3² + …. + 3^k-1 = (3^k – 1)/2  ...(i)

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

consider 

1 + 3 + 3² + ….+ 3^k-1 + (3^(k+1)-1  

= (1 + 3 + 3² + .... + 3^{k - 1}) + 3^k 

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.

2. Prove the following by using the principle of mathematical induction for all n ∈ N: 1³ + 2³ + 3³+ ….+ n³ = [n(n+1)/2]² 

Answer :

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

P(n) : 1³ + 2³ + 3³ + …. + n³ = [n(n+1)/2]²  

For n = 1, we have 

3. Prove that following by using the principle of mathematical induction for all n ∈ N :  1 + 1/(1 + 2) + 1/(1 + 2 + 3) + .... + 1/(1 + 2 + 3 + ...n) = 2n/(n + 1) 

Answer :

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

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. 

4. 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) = [n(n + 1)(n + 2)(n +3)]/4 

Answer :

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

P(n) : 1.2.3 + 2.3.4 + .... + n(n + 1)(n+2) = [n(n+1)(n+2)(n+3)]/4 

For n = 1 , we have 

P(1) : 1.2.3 = 6 = [1(1 + 1)(1 + 2)(1 + 3)]/4 = (1.2.3.4)/4 = 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) = [k(k+1)(k + 2)(k + 3)]/4 ...(i)

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 .

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

5. Prove the following by using the principle of mathematical induction for all n ∈ N: 1.3 + 2.3² + 3.3³ + .... + n.3ⁿ =[(2n - 1)3ⁿ⁺¹ + 3]/4 

Answer :

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

P(n) : 1.3 + 2.3² + 3.3³ + .... + n.3ⁿ  =[(2n - 1)3ⁿ⁺¹ + 3]/4 

For n = 1, we have  

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

1.3 + 2.3² + 3.3³ + .... + k.3^k[(2k-1)3^k+1 + 3]/4 ...(i) 

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

Consider,

1.3 + 2.3² + 3.3³ + .... + k3^k = [(2k -1)3^k+1 + 3]/4 ...(i) 

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

Consider,

1.3 + 2.3² + 3.3³ + ...+k3^k + (k +1)3^k+1 

= (1.3 + 2.3² + 3.3³ + .... + k.3^k ) + (k + 1)3^k+1 

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.

6. Prove the following by using the principle of mathematical induction for all n ∈ N: 1.2 + 2.3+ 3.4 + ....+ n.(n + 1) = [{n(n+1)(n+2)}/3] 

Answer :

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

P(n) : 1.2 + 2.3+ 3.4 + .....+ n.(n + 1) = [{n(n+1)(n+2)}/3] 

For n = 1, we have 

P(1) : 1.2 = 2 = [1(1 +1)(1 + 2)]/3 = 1.2.3/3 = 2 , which is true.  

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

1.2 + 2.3 + 3.4 + .... + k.(k + 1) = [{k(k+1)(k+2)}/3] ...(i) 

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

Consider  

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

= [1.2 + 2.3 + 3.4 + ..... + k.(k + 1)] + (k + 1).(k + 2)

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.

7. Prove the following by using the principle of mathematical induction for all n ∈ N:  1.3 + 3.5 + 5.7 + .... + (2n - 1)(2n + 1) = [n(4n² + 6n - 1)]/3 

Answer :

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

P(n): 1.3+ 3.5 + 5.7 + ....+(2n - 1)(2n + 1) = [n(4n² + 6n - 1)]/3 

For n = 1, we have 

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.

8. 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.2^k = (k – 1) 2^{k + 1} + 2 …(i)

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

Consider

{1.2 + 2.2² + 3.2³ + ..... +k.2^k } + {k + 1}. 2^k+1 

= (k - 1) 2^k+1 + 2 + (k + 1)2^k+1 

= 2^k+1 {(k - 1) + (k + 1)} + 2 

= 2^k+1  . 2k + 2 

= k.2^{(k+1) + 1}  +  2 

= {(k + 1) - 1}2^(k+1)+1  +  2  

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.

9. Prove the following by using the principle of mathematical induction for all n ∈ N : 1/2 + 1/4 + 1/8 + .... + 1/2ⁿ = 1 - 1/2ⁿ 

Answer :

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

P(n) : 1/2 + 1/4 + 1/8 + .... + 1/2ⁿ  = 1 - 1/2ⁿ 

For n = 1, we have  

P(1) : 1/2 = 1 - 1/2¹  = 1/2 , which is true.  

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

1/2 + 1/4 + 1/8 + ....+ 1/2^k  = 1 - 1/2^k  ...(i) 

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

10. Prove the following by using the principle of mathematical induction for all n ∈ N:  1/2.5 + 1/5.8 + 1/8.11 + .... + 1/[(3n - 1)(3n + 2)] = n/(6n + 4)

Answer :

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

P(n) : 1/2.5 + 1/5.8 + 1/8.11 + .... + 1/[(3n - 1)(3n + 2)] = n/(6n + 4) 

For n = 1, we have  

P(1) = 1/2.5 = 1/10 = 1/(6.1 + 4) = 1/10, which is true.  

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

1/2.5 + 1/5.8 + 1/8.11 + .... + 1/[(3k - 1)(3k+ 2)] = k/(6k + 4)  ...(i) 

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

11. Prove the following by using the principle of mathematical induction for all n ∈ N: 1/(1.2.3) + 1/(2.3.4) + 1/(3.4.5) + ....+ 1/[n(n+1)(n+2)] = n(n+3)/[4(n+1)(n+2)].

Answer :

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

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.

12. Prove the following by using the principle of mathematical induction for all n ∈ N: a + ar + ar² + .....+ ar^n-1 = a(rⁿ - 1)/(r - 1) 

Answer :

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

P(n): a + ar + ar² + .....+ ar^n-1 = a(rⁿ - 1)/(r - 1) 

For n = 1, we have 

P(1): a =   a(r¹ - 1)/(r - 1)  = a , which is true.  

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

a + ar + ar² + ....+ ar^k-1 = a(r^k - 1)/(r - 1)  ...(i) 

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

Consider  

{a + ar + ar² + .....+ ar^k-1 } + ar^(k+1)-1 

= a(r^k - 1)/(r - 1) + ar^k  [Using (i)] 

 

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

(1+3/1)(1+ 5/4)(1+7/9)...[1 + (2n+1)/n² ) = (n+ 1)²  

Answer :

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

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.

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

(1 + 1/1)(1 + 1/2)(1+ 1/3).....(1 + 1/n) = (n + 1) 

Answer :

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

P(n) : (1 + 1/1)(1 + 1/2)(1+ 1/3).....(1 + 1/n) = (n + 1)  

For n = 1, we have 

P(1): (1 + 1/1) = 2 = (1 + 1) , which is true.  

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

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.

15. Prove the following by using the principle of mathematical induction for all n ∈ N: 
1² + 3² + 5² + ... + (2n - 1)² = [n(2n - 1)(2n+1)]/3 

Answer :

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

P(n) = 1² + 3² + 5² + ... + (2n - 1)² = [n(2n - 1)(2n+1)]/3 

For n = 1, we have 

P(1) = 1² = 1 = [1(2.1 - 1)(2.1 + 1)]/3 = 1.1.3/3 = 1, which is true. 

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

P(k) = 1² + 3² + 5² + ...+(2k - 1)²  = [k(2k-1)(2k+1)]/3 ...(1) 

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

16. Prove the following by using the principle of mathematical induction for all n ∈ N: 
1/1.4 + 1/4.7 + 1/7.10 + ... + 1/[(3n-2)(3n+1)] = n/(3n+1) 

Answer :

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

P(n) : 1/1.4 + 1/4.7 + 1/7.10 + ... + 1/[(3n-2)(3n+1)] = n/(3n+1)  

For n = 1, we have 

P(1) = 1/1.4 = 1/(3.1 + 1) = 1/4 = 1/1.4 , which is true.  

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

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.

17. Prove the following by using the principle of mathematical induction for all n ∈ N: 
1/3.5 + 1/5.7 + 1/7.9 + ....+ 1/[(2n+1)(2n+3)] = n/3(2n+3) 

Answer :

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

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.

18. Prove the following by using the principle of mathematical induction for all n ∈ N: 1+ 2 + 3 + ... + n <(1/8)(2n + 1)².

Answer :

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

1+2+....+ k < (1/8)(2n + 1)²

Adding (k + 1) on both the sides of the inequality , we have,  

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.

19. 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

(k + 1){(k+1) + 1}{(k+1) + 5} 

= (k+1)(k+2){(k+5) + 1}

= (k + 1)(k + 2)(k+5)+(k+1)(k+2)

= {k(k+1)(k+5)+2(k+1)(k+5)}+ (k+1)(k+2) 

= 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 

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

20. Prove the following by using the principle of mathematical induction for all n ∈ N: 10^2n–1 + 1 is divisible by 11.

Answer :

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

P(n): 10^2n–1 + 1 is divisible by 11.

It can be observed that P(n) is true for n = 1 since P(1) = 10^2.1–1 + 1 = 11, which is divisible by 11.

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

10^2k–1 + 1 is divisible by 11.

∴10^2k–1 + 1 = 11m, where m ∈ N …(1)

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

Consider

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 

= 10² .11 m - 100 + 1 [Using (1)]

= 100 × 11m - 99 

= 11(100m - 9) 

= 11r, where r = (100m - 9) is some natural number 

Therefore, 10^2(k+1)-1 + 1 is divisible by 11. 

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.

21. 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 x^2 × 1 – y^2 × 1 = x² – y² = (x + y) (x – y) is divisible by (x + y).

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

x^2k – y^2k is divisible by x + y.

∴x^2k – y^2k = 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. 

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

22. 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 3^{2 × 1 + 2} – 8 × 1 – 9 = 64, which is divisible by 8.

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

3^2k+2 – 8k – 9 is divisible by 8.

∴3^2k+2 – 8k – 9 = 8m; where m ∈ N …(1)

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

Consider 

3^2(k+1)+2 - 8(k+1) - 9 

= 3^2k+2.3²  -8k-8-9 

= 3² (3^2k+2  -8k - 9 + 8k + 9) - 8k - 17 

= 3² (3^2k+2 - 8k - 9) + 3² (8k + 9) - 8k - 17 

= 9.8m + 9(8k + 9) - 8k - 17 

= 9.8m + 72k + 81 - 8k - 17 

= 9.8m + 64k + 64

= 8(9m + 8k + 8)

= 8r, where r = (9m + 8k + 8) is a natural number 

Therefore, 3^2(k+1)+2 - 8(k+1) - 9 is divisible by 8. 

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

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

23. 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 41¹ - 14¹ = 27, which is a multiple of 27.

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

41^k – 14^kis a multiple of 27

∴41^k – 14^k = 27m, where m ∈ N …(1)

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

Consider

41^k+1 - 14^k+1 

= 41^k . 41- 14^k .14 

= 41(41^k - 14^k + 14^k ) - 14^k .14 

= 41(41^k - 14^k ) + 41.14^k - 14^k .14 

= 41.27m + 14^k (41-14)

= 41.27m + 27.14^k 

= 27(41m - 14^k ) 

= 27× r, where r = (41m - 14^k ) is a natural number 

Therefore, 41^k+1 - 14^k+1 is a multiple of 27.

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.

24. Prove the following by using the principle of mathematical induction for all n ∈ N (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

[2(k+1)+7] = (2k + 7) + 2 

∴ [2(k+1)+7] = (2k+7)+2 <(k+3)² + 2  [using (1)] 

⇒ 2(k+1) + 7 < k² + 6k + 9 + 2 

⇒ 2(k+1) + 7 < k² + 6k + 11 

Now, k² + 6k + 11 < 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, by the principle of mathematical induction, statement P(n) is true for all natural number i.e., n.