The problem:

Define the recurrence relation $\displaystyle H(n)$ by the rules :

$\displaystyle H(0) = 3 , H(1) = -3, H(n) = 2H(n-1) - H(n-2) + 6$ for $\displaystyle n>=2$. Calculate $\displaystyle H(6)$. Prove by induction that $\displaystyle H(n)$ is divisible by 3 for all $\displaystyle n$ (n - positive integer).

I calculate $\displaystyle H(6)$ using simple algebra and get $\displaystyle H(6) = 57$ (I will be grateful if you can check that, so I didn't miss anything). But I'm stuck on the proving by induction part. I need to show somehow that sum of the digits of the number $\displaystyle H(n)$ are divisble by 3, but I don't know how to do it. I will appreciate any ideas or solutions, thank you