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

Using the formula for the sum of a geometric series:
$$
\sum_{k=0}^\infty x^k=\frac1{1-x}\tag{1}
$$
Taking the derivative of $(1)$ and tossing the terms which are $0$:
$$
\sum_{k=1}^\infty kx^{k-1}=\frac1{(1-x)^2}\tag{2}
$$
Taking the derivative of $(2)$ and tossing the terms which are $0$:
$$
\sum_{k=2}^\infty k(k-1)x^{k-2}=\frac2{(1-x)^3}\tag{3}
$$
Reindexing the sum in $(3)$:
$$
\sum_{k=0}^\infty(k+2)(k+1)x^k=\frac2{(1-x)^3}\tag{4}
$$
Plug in $x=\frac14$:
$$
\begin{align}
\sum_{k=0}^\infty\frac{(k+2)(k+1)}{4^k}
&=\frac2{\left(\frac34\right)^3}\\[6pt]
&=\frac{128}{27}\tag{5}
\end{align}
$$
