Q.Prove that $4^n - 1$ is divisible by $3$ for every natural number $n$.

$2^{3n}-7n-1$ is divisible by $49$ for every $n\in N$ (provable by induction); since $49=7^2$, it is in particular always divisible by $7$, which is the matching option here.

Let $P(n): 2^{3n}-7n-1$ is divisible by $49$.

Base case ($n=1$): $2^3-7(1)-1 = 8-7-1=0$, which is divisible by $49$.

Inductive step: Assume $P(k)$ holds, i.e. $2^{3k}-7k-1=49m$ for some integer $m$. Then

$2^{3(k+1)}-7(k+1)-1 = 8\cdot2^{3k}-7k-8$

$= 8(2^{3k}-7k-1) + 56k+8-7k-8$

$= 8(49m) + 49k = 392m+49k = 49(8m+k)$

which is divisible by $49$. So $P(k+1)$ holds, and by induction $P(n)$ holds for all $n\in N$.

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$7^n-3^n$ is divisible by $4$; option (c).

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