Proof $\sum_{n=0}^\infty \frac{n^2+3n+2}{4^n} = \frac{128}{27}$ 
$\sum_{n=0}^\infty \frac{n^2+3n+2}{4^n} =  \frac{128}{27}$ Given hint: $(n^2+3n+2) = (n+2)(n+1)$

I've tried converting the series to a geometric one but failed with that approach and don't know other methods for normal series that help determine the actual convergence value. Help and hints are both appreciated

$$
\begin{aligned}
& S_{0} = \sum_{n=0}^{\infty} \frac{n^{0}}{4^{n}} = \sum_{n=0}^{\infty} \frac{1}{4^{n}} = \sum_{n=0}^{\infty} (1/4)^{n} = \frac{1}{1 - (1/4)} \Rightarrow \color{red}{S_{0} = \frac{4}{3}} \\ \\
& S_{1} = \sum_{n=0}^{\infty} \frac{n^{1}}{4^{n}} = \sum_{n=0}^{\infty} \frac{n + 1 - 1}{4^{n}} = \sum_{n=0}^{\infty} \frac{n + 1}{4^{n}} - \sum_{n=0}^{\infty} \frac{1}{4^{n}} = 4 \sum_{n=0}^{\infty} \frac{n + 1}{4^{n + 1}} - S_{0} \\
& \qquad = 4 \sum_{n=1}^{\infty} \frac{n}{4^{n}} - S_{0} = 4 \left[ - 0 + \sum_{n=0}^{\infty} \frac{n}{4^{n}} \right] = 4 S_{1} - S_{0} \Rightarrow \color{red}{S_{1} = \frac{1}{3} S_{0} = \frac{4}{9}} \\ \\
& S_{2} = \sum_{n=0}^{\infty} \frac{n^{2}}{4^{n}} = \sum_{n=0}^{\infty} \frac{(n + 1)^{2} - 2 n - 1}{4^{n}} = 4 \sum_{n=0}^{\infty} \frac{(n + 1)^{2}}{4^{n+1}} - 2 \sum_{n=0}^{\infty} \frac{n}{4^{n}} - \sum_{n=0}^{\infty} \frac{1}{4^{n}} \\
& \qquad = 4 S_{2} - 2 S_{1} - S_{0} = 4 S_{2} - \frac{5}{3} S_{0} \Rightarrow \color{red}{S_{2} = \frac{5}{9} S_{0} = \frac{20}{27}} \\ \\
& \sum_{n=0}^{\infty} \frac{n^{2} + 3 n + 2}{4^{n}} = S_{2} + 3 S_{1} + 2 S_{0} = \frac{20}{27} +  \frac{12}{9} +  \frac{8}{3} = \frac{20 + 36 + 72}{27} = \frac{128}{27} \\ \\
\end{aligned}
$$
