<s>[INST] 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 
 [/INST] *

*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 $n\ge 1$
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