If $n$ is a positive integer, then $2 \cdot 4^{2 n+1}+3^{3 n+1}$ is divisible by
If $n$ is a positive integer, then $2 \cdot 4^{2 n+1}+3^{3 n+1}$ is divisible by
$2$
$9$
$11$
$27$
Solution
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 .