Show that $7\mid(3^{2n+1}+2^{n+2})$ for all $n\in\mathbb{N}$ 
Prove that the following is true for every $n∈ℕ$:
$$7\mid(3^{2n+1}+2^{n+2}).$$

I've noticed
$$3^{2n+1}+2^{n+2} =3^{2n} \cdot 3+2^{n} \cdot 4.$$
Any suggestions how to continue from there to get something like $7k$ for $k\in\mathbb{N}$.
Thank you in advance!

\begin{align}3^{2n+1}+2^{n+2}=3\cdot 9^n + 4\cdot 2^n&=7\cdot 9^n -4(9^n-2^n)\\
&=7\cdot 9^n-4(9-2)(9^{n-1}+9^{n-2}\cdot2+\cdots+2^{n-1})\\
&=7[ 9^n-4(9^{n-1}+9^{n-2}\cdot 2\cdots+2^{n-1})]\end{align}
Which is divisible by $7$
Here is a solution based on congruences,
$$3\cdot 9^n +4\cdot 2^n \equiv 3\cdot 2^n +4\cdot 2^n \equiv 7\cdot 2^n \equiv 0\text{(mod 7)}$$
