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 $P(k) \equiv 2^{3k} - 7k - 1 = 49I$
$\therefore P(k + 1) \equiv 2^{3k + 3} - 7k - 8$
$= 8(49I + 7k + 1) - 7k - 8$
$= 49(8I) + 49k = 49I_{1}$
Hence, by mathematical induction $2^{3n} - 7n - 1$ is divisible by 49.