Find a generating function for the recurrence relation given by $b_{n+1} = 3b_{n} + 2^{n}$, $b_{0} = 1$.

Question

I've gotten a bit stuck.. Let $G(x) = \sum_{0}^{\infty} b_{n} x^{n}$. Now: $\sum_{0}^{\infty} b_{n+1} x^{n+1} = \sum_{0}^{\infty} 3 b_{n} + \sum_{0}^{\infty} 2^n x^{n+1}$ yields $G(x) - b_{0} = 3xG(x) + x \sum_{0}^{\infty} 2^{n}x^{n}$.

Not sure how to get this last sum in a closed form to continue the problem