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.,
23k – 7k – 1 is divisible by 49, i.e.,
23k – 7k – 1 = 49m, m∈N
Now 23(k + 1) – 7 (k + 1) – 1 = 23k . 23 – 7k – 7 – 1
= 8.23k – 7k – 8 = 8 (23k – 7k – 1) + 49k
= 8. 49m + 49k = 49 (8m + k)
⇒ 23(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.