Use Principle of Mathematical Induction to show that for all $n\:\in \mathbb{N}$, $a_n=2^{n+2}\cdot 5^{2n+1}+3^{n+2}\cdot 2^{2n+1}$ is divisible by 19 Here's what I did:
Base Case: $n = 1$, $2^3*5^3+3^3*2^3$ = $1216$ = $19(64)$
Induction Hypothesis (IH): $a_k=2^{k+2}\cdot 5^{2k+1}+3^{k+2}\cdot 2^{2k+1}$ for some $k\:\in \mathbb{N}$.
Inductive Step: $a_{k+1}=2^{\left(k+1\right)+2}\cdot \:5^{2\left(k+1\right)+1}+3^{\left(k+1\right)+2}\cdot \:2^{2\left(k+1\right)+1}$
$a_{k+1}=2^{\left(k+2\right)}\cdot 2\cdot \:5^{2k+1}\cdot 5^2+3^{k+2}\cdot 3\cdot \:2^{2k+1}\cdot 2^2$
$a_{k+1}=2^{k+2}\:\cdot 5^{2k+1}\cdot 50+3^{k+2}\cdot 2^{2k+1}\cdot 12$
How do I prove that the latter statement is divisible by 19?
(I've tried to set up the equation in a way where I know that $2^{k+2}\:\cdot 5^{2k+1}$ and $3^{k+2}\cdot 2^{2k+1}$ is divisible by 19. I'm not sure about the 50 and the 12.)

You're almost there! Hint for the finish:
$$
2^{k+2}\cdot 5^{2k+1}\cdot 50+3^{k+2}\cdot 2^{2k+1}\cdot 12\\
= 12\left(2^{k+2}\cdot 5^{2k+1}+3^{k+2}\cdot 2^{2k+1}\right) + 38\left(2^{k+2}\cdot 5^{2k+1}\right)
$$
