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.

You can use modular arithmetic :
Suppose $3^{3n+1}+2^{n+1}$ is divisible by 5, then
$3^{3n+1} \equiv -2^{n+1}[5]$
so you have 
$$3^{3(n+1)+1}+2^{n+1+1} \equiv 27 \times 3^{3n+1}+2 \times2^{n+1} [5]$$
$$3^{3(n+1)+1}+2^{n+1+1} \equiv 27 \left(  -2^{n+1} \right) +2 \times2^{n+1} [5]$$
$$3^{3(n+1)+1}+2^{n+1+1} \equiv -25 \times2^{n+1} [5]$$
$$3^{3(n+1)+1}+2^{n+1+1} \equiv 0 [5]$$
