Prove $3^{2n+1} + 2^{n+2} = 7m$ I'm trying to prove that $3^{2n+1} + 2^{n+2}$ is a multiple of 7 by using induction. 
So I started to prove it for $n=1$: $3^{2(1)+1}+2^{1+2}=3^3+2^3=27+8=35=7(5)$. 
Next, try to prove that the statement being true $n=k$ implies it being true for $n=k+1$. Thus: 
$3^{2(k+1)+1}+2^{(k+1)+2} = 3^{2k+3}+2^{k+3} = (3)(3)(3^{2k+1})+(2)(2^{k+2}) = 9(3^{2k+1})+2(2^{k+2})$
I feel like I'm almost there, if I could've factor 9 and 2 somehow I could say that $3^{2k+1}+2^{k+2}=7m$ for some integer $m$, but I can't find a way to do it. What am I missing? Or did I do a blunder somewhere along the road?
Thanks in advance.

For a non-inductive solution, note that $3^2\equiv 2 \bmod 7$; then
$\begin{align}
3^{2n+1} + 2^{n+2} &\equiv 3\cdot 2^n + 4\cdot 2^n \\
&\equiv 7\cdot 2^n \bmod 7\\
\end{align}$
