Q.Prove that $n(n+1)(2n+1)$ is divisible by $6$ 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 $n\in N$.

Checking directly: $n=2$ gives $2^6-14-1=64-15=49$; $n=3$ gives $2^9-21-1=512-22=490=49\times10$ — both divisible by $49$, and hence also by $7$ (since $49=7\times7$). Among the given options $\{14,121,7,98\}$, only $7$ divides every one of these values (e.g. $49$ is not divisible by $14$, $98$, or $121$), so $7$ is the correct choice from this list, even though the sharper true result is divisibility by $49$.

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

Since $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\dots+b^{n-1})$, the factor $a-b$ always divides $a^n-b^n$. With $a=7,b=3$, $a-b=4$, so $4\mid 7^n-3^n$. (This can also be proved by induction — see Q73.)