$\begin{aligned}
& \text { Let } P(n)=2^{3 n}-7 n-1 \\
& \Rightarrow \quad P(1)=2^{3(1)}-7(1)-1=8-8=0 \\
& \Rightarrow \quad P(2)=2^{3(2)}-7(2)-1=64-15=49 \\
&
\end{aligned}$
$P(1)$ and $P(2)$ are divisible by 49 .
Let $P(k)=2^{3 k}-7 k-1=49 t$, where $t$ is an integer
Now,
$\begin{aligned}
P(k+1) & =2^{3(k+1)}-7(k+1)-1=2^{3 k} \cdot 2^3-7 k-7-1 \\
& =8\left(2^{3 k}-7 k-1\right)+49 k \\
& =8(49 t)+49 k \\
& =49(8 t+k), \text { where } 8 t+k \text { is an integer }
\end{aligned}$
Thus, $2^{3 n}-7 n-1$ is divisible by 49 .