Prove by induction that $3^{3n+1} + 2^{n+1}$ is divisible by 5 How do I do this? I've tried using logarithms, factoring, but nothing seems to work.

Other users have already outlined the proof by induction, but I think a direct proof is interesting as well. 
By Fermat's little theorem (or by inspection), we know that 
$$3^4 \equiv 2^4 \equiv 1 \pmod{5}$$
This means that powers of $3$, modulo $5$, are determined by the exponents remainder when divided by $4$.
$$3^{3n+1} \equiv \begin{cases}3 \pmod{5} &\mbox{if }\,\, n \equiv 0 \pmod{4}\\
1 \pmod{5} &\mbox{if }\,\, n \equiv 1 \pmod{4}\\
2 \pmod{5} &\mbox{if }\,\, n \equiv 2 \pmod{4}\\
4 \pmod{5} &\mbox{if }\,\, n \equiv 3 \pmod{4}\\
\end{cases}$$
$$2^{n+1} \equiv \begin{cases}2 \pmod{5} &\mbox{if }\,\, n \equiv 0 \pmod{4}\\
4 \pmod{5} &\mbox{if }\,\, n \equiv 1 \pmod{4}\\
3 \pmod{5} &\mbox{if }\,\, n \equiv 2 \pmod{4}\\
1 \pmod{5} &\mbox{if }\,\, n \equiv 3 \pmod{4}\\
\end{cases}$$
For each possible residue for $n$ modulo $4$, the sum is $0$ modulo $5$. 
