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!
 A: Note that $2 \equiv 3^2 \pmod 7$, therefore
$$
3^{2n+1} + 2^{n+2} \equiv 3^{2n+1} + 3^{2n+4} \equiv 3^{2n+1} \cdot 28 \pmod 7,
$$
and that is of course divisible by $7$.
A: \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,
$$3\cdot 9^n +4\cdot 2^n \equiv 3\cdot 2^n +4\cdot 2^n \equiv 7\cdot 2^n \equiv 0\text{(mod 7)}$$
A: We have
$$3^{2n+1}+2^{n+2}=3\times\color{red}9^n+4\times2^n\equiv3\times\color{red}2^n+4\times2^n=7\times2^n\equiv0\mod7$$
A: Note that a sequence with form $u_n=a\cdot 9^n+b\cdot 2^n$ satisfies a linear recurrence with auxiliary equation $$(x-9)(x-2)=0$$ so that $$u_{n+1}=11u_n-18u_{n-1}$$
Hence if $7|u_n \& 7|u_{n-1}$ then $7|u_{n+1}$. We have $u_0=7, u_1=35$ both divisible by $7$, so with a base case established we are done by induction.
If $u_n=7k_n$, then $k_n$ satisfies the same recurrence with $k_0=1, k_1=5$
A: 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 $7$.
