Find all $n$ such that $3^{2n+1}+2^{n+2}$ is divisible by $7$

Note that

$$\begin{split} 3^{2(k+1)+1} + 2^{(k+1)+2} &= 9\cdot 3^{2k+1} + 2\cdot 2^{k+2}\\ &= 7 \cdot 3^{2k+1} + 2 (3^{2k+1} + 2^{k+2}). \end{split}$$

Let $A_n=3^{2n+1}+2^{n+2}$ then you will find that $A_{n+1}=11A_n-18A_{n-1}$

Rationale: If $u_n=A\alpha^n+B\beta^n$ it is easy to check that $u_{n+1}=(\alpha+\beta)u_n-\alpha\beta u_{n-1}$.

Set $\alpha = 3^2=9, \beta=2$.

You need two consecutive values to ensure the persistence of the factor $7$ (you could use $n=-1$ even though the value involves fractions), which makes it less attractive in some ways than the induction arguments with a single base case. However, this can also used to construct further examples of persistence, and is quick if you are doing multiple questions of the same type.