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!

Just surprised that the very crude induction method with $u_n=3\cdot 9^n+4\cdot 2^n$ hasn't been posted yet - it goes $$u_{n+1} = 3\cdot 9^{n+1}+4\cdot 2^{n+1}=7\cdot3\cdot 9^n+2\cdot 3\cdot 9^n+2\cdot 4\cdot 2^n=7\cdot 3\cdot 9^n+2\cdot u_n$$ which is divisible by $7$ if $u_n$ is divisible by $7$. Base case is $u_0=7$. It depends on the fact that $9-2$ is divisible by $7$.
