Question

For each n∈N, 49ⁿ + 16n – 1 is divisible by

(a) 3 

(b) 29 

(c) 19 

(d) 64

Answer 1

Answer:(D) = 64

For n = 1, 49¹ + 16 × 1 – 1 = 49 + 15 = 64 

∴ For n = 1, 49ⁿ + 16n – 1 is divisible by 64 and not by any of the other given alternatives. 

∴ We shall prove using mathematical induction, that 49ⁿ + 16n – 1 is divisible by 64 ∀ n∈N.

 Let T(n) be the statement: 49ⁿ + 16n – 1 is divisible by 64 

Basic Step: 

For n = 1, T(1) is divisible by 64 as proved above. 

Induction Step: 

Assume T(k) to be true i.e., 

T(k) = 49^k + 16k – 1 is divisible by 64, i.e., 

49^k + 16k – 1 = 64m, m∈N. ...(i) 

∴ T(k + 1) = 49^{k + 1} + 16(k + 1) – 1 

= 49. 49^k + 16k + 16 – 1 

= 49. 49^k + 16k + 15 

= 49(49^k + 16k – 1) – 48(16k) + 64 

= 49 (64m) – 12 (64k) + 64 

= 64 (49m – 12k + 1) 

⇒ 49^{k + 1} + 16 (k + 1) – 1 is divisible by 64. 

⇒ T(k + 1) is true whenever T(k) is true. 

⇒ 49ⁿ + 16n – 1 is divisible by 64 ∀  n∈N