Find all $n$ such that $3^{2n+1}+2^{n+2}$ is divisible by $7$ Find all $n$ such that 
$3^{2n+1}+2^{n+2}$ is divisible by $7$
Prove that your answer is correct 
So I am not allowed to use mods,  as is a calculus question, I have tried by induction but can't get to prove that it works for $k+1$, by multiplying the equation by powers of $2$ and $3$. 
Thank you for your help 

Just another way: 
Inductive hypothesis: $3^{2n+1} = 7k - 2^{n+2}$
Inductive step:
$$3^{2n+3} + 2^{n+3} = 3^2 * 3^{2n+1} + 2 * 2^{n+2} $$
$=3^2(7k - 2^{n+2})+2*2^{n+2}$
$=7k(3^2) - 3^2*2^{n+2} + 2* 2^{n+2}$
$=7k(3^2) + 2^{n+2}(-3^2+2)$
$=7k(3^2) -7*(2^{n+2})$
Both terms are divisible by 7 
