Prove that $7 | (3^{2n + 1} + 2^{n +2})$ Prove that $7 | (3^{2n + 1} + 2^{n +2})$
So far I have:
Base case:  n = 1
$ = (3^{2(1) + 1} + 2^{(1) +2})$
$ = (3^{3} + 2^{3})$
$ = (35)$ which divides 7
Inductive Step:
$ = (3^{2(n +1) + 1} + 2^{(n +1) +2})$
$ = (3^{2n + 3} + 2^{n + 3})$
$ 7j = (3^{2n + 3} + 2^{n + 3})$
-I'm a bit stuck as to how to proceed, any help or correcitons are appreciated.
- I believe I understand where we need to go now but i'm not sure how to get there. It seems obvious that any n plugged into this formula would be divisible by 7 given that $3^1 + 2^2 = 7$ Since you have that $3^1+2^2$ nested inside the given formula it seems easily understood that any sum with any positive n would also be divisible by 7. I just don't see the steps in between to actually prove it.

Hint: if you want to use induction:
if $7\mid 3^{2k+1}+2^{k+2}=3\cdot 9^k+4\cdot 2^k$ for some $k\in\mathbb Z_{\ge 0}$, then:
$$3\cdot 9^{k+1}+4\cdot 2^{k+1}=2\left(3\cdot 9^k+4\cdot 2^k\right)+7\left(3\cdot 9^k\right)$$
Otherwise, if you know Modular Arithmetic:
$$3\cdot 9^n+4\cdot 2^n\equiv 3\cdot 2^n+4\cdot 2^n\equiv 7\cdot 2^n\equiv 0\pmod{7}$$
