Question

For all n∈N, 2³ⁿ – 7n – 1 is divisible by

(a) 64 

(b) 36 

(c) 49 

(d) 25

Answer 1

Answer: (C) = 49

For n = 1, 

2³ⁿ – 7n – 1 = 2³ – 7 – 1 = 8 – 8 = 0 which is divisible by all the given alternatives. 

For n = 2, 

2³ⁿ – 7n – 1 = 2⁶ – 7 × 2 – 1 = 64 – 14 – 1 = 49, which is divisible by only 49 out of the given alternatives. 

∴ We need to prove 2³ⁿ – 7n – 1 is divisible by 49 ∀  n∈N. 

Let T (n) be the statement:

2³ⁿ – 7n – 1 is divisible by 49 

Basic Step: 

For n = 1, 2³ⁿ – 7n – 1 = 0, divisible by 49 

⇒ T(1) is true. 

Induction Step: 

Assume T(k) is true ∀  k∈N, i.e., 

2^3k – 7k – 1 is divisible by 49, i.e.,

2^3k – 7k – 1 = 49m, m∈N

Now 2^{3(k + 1)} – 7 (k + 1) – 1 = 2^3k . 2³ – 7k – 7 – 1

 = 8.2^3k – 7k – 8 = 8 (2^3k – 7k – 1) + 49k 

= 8. 49m + 49k = 49 (8m + k) 

⇒ 2^{3(k + 1)} – 7 (k + 1) – 1 is divisible by 49 

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

⇒ 2³ⁿ – 7n – 1 is divisible by 49 ∀  n∈N.