If $n$ is a positive integer, then $2 \cdot 4^{2 n+1}+3^{3 n+1}$ is divisible by

Given expression, $p(n)=2 \cdot 4^{2 n+1}+3^{3 n+1}$ Let $n=1$ $p(1)=2 \cdot 4^{2+1}+3^{3+1}=209$ which is divisible by 11 . Let $p(k)$ is also divisible by 11 . $p(k)=2 \cdot 4^{2 k+1}+3^{3 k+1}$ is divisibly by 11 ...(i) Now we will prove that $p(k+1)$ is true. $p(k+1)=2 \cdot 4^{2(k+1)+1}+3^{3(k+1)+1}$ $\begin{aligned} & =2 \cdot 4^{2 k+1} \cdot 16+3^{3 k+1} \cdot 27 \\ & =2 \cdot 4^{2 k+1} \cdot 16+3^{3 k+1} \cdot 16+3^{3 k+1} \cdot 11 \\ & =16\left(2 \cdot 4^{2 k+1}+3^{3 k+1}\right)+3^{3 k+1} \cdot 11\end{aligned}$ Divisible by 11 from Eq. (i) $\therefore p(k+1)$ is also true. $\therefore p(n)$ is divisible by 11 .