Prove by induction: $2^{n+2} +3^{2n+1}$ is divisible by $7$ for all $n \in \mathbb{N}$ I want to prove that $2^{n+2} +3^{2n+1}$ is divisible by $7$ for all $n \in \mathbb{N}$ using proof by induction.
Attempt
Prove true for $n = 1$
$2^{1+2} + 3^{2(1) +1} = 35$
35 is divisible by 7 so true for $n =1$
Induction step: Assume true for $n = k$ and prove true for $n = k+1$
$n = k$
$2^{k+2} + 3^{2k +1} = 2^k \cdot 2^2 + 3^{2k} \cdot 3^1 = (4) 2^k + (3) 3^{2k}$
$n = k+1$
$2^{k+3} +3^{2k+3} = (8) 2^k + (27) 3^{2k}$
Thoughts:
I know I need to use the expression for $n = k$ to prove that it is true for $n = k+1$ but I am not sure where to start. I think that using modular arithmetic would be too complex.  Any ideas would be greatly appreciated!
Update
Thanks for all of the useful answers/suggestions. I understand the approach to take now!

\begin{eqnarray*}
2^{k+3}+3^{2k+3} =(9-7)2^{k+2} +9 \times 3^{2k+1}= 9(\color{blue}{2^{k+2}+3^{2k+1}})-\color{blue}{7} \times2^{k+2}
\end{eqnarray*}
