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
Note that 

$$\begin{split}
3^{2(k+1)+1} + 2^{(k+1)+2} &amp;= 9\cdot 3^{2k+1} + 2\cdot 2^{k+2}\\
&amp;= 7 \cdot 3^{2k+1} + 2 (3^{2k+1} + 2^{k+2}). 
\end{split}$$
Setting $n=1$, we get $$3^{2\cdot 1+1}+2^{1+2}=35$$ above number $35$ is divisible by $7$ hence it holds for $n=1$
Assume the number $3^{2n+1}+2^{n+2}$ is divisible by $7$ for $n=k$ then $$3^{2k+1}+2^{k+2}=7\lambda \tag 1$$
Setting $n=k+1$, we get $$3^{2k+2+1}+2^{k+1+2}$$
$$=9\cdot 3^{2k+1}+2\cdot 2^{k+1}$$
$$=9\cdot 3^{2k+1}+9\cdot 2^{k+1}-7\cdot 2^{k+1}$$
$$=9(3^{2k+1}+2^{k+1})-7\cdot 2^{k+1}$$
setting the value from (1), 
$$=9(7\lambda)-7\cdot 2^{k+1}$$
$$=7(9\lambda-2^{k+1})$$ 


since, $(9\lambda-2^{k+1})$ is some integer hence the number $7(9\lambda-2^{k+1})$ is divisible by $7$ 

hence the number $(3^{2n+1}+2^{n+1})$ is divisible by $7$ for all integers $n\ge 1$