Explanation

(c)
Let $P(n) = 2^{3n} - 7n - 1$
$\therefore\ \ \ \ \ \ \ \ \ \ P(1) = 0,\ P(2) = 49$
$P(1)$ and $P(2)$ are divisible by 49.
Let $P(k) \equiv 2^{3k} - 7k - 1 = 49I$
$\therefore P(k + 1) \equiv 2^{3k + 3} - 7k - 8$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 8(49I + 7k + 1) - 7k - 8$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 49(8I) + 49k = 49I_{1}$
Alternate
$\ \ \ \ \ \ \ \ \ \ P(n) = (1 + 7)^{n} - 7n - 1$
$= 1 + 7n + 7^{2}\frac{n(n - 1)}{2!} + \ldots - 7n - 1$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 7^{2}\left( \frac{n(n - 1)}{2!} + \ldots \right)$