<s>[INST] 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
 [/INST] 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 $x=\frac14$:
$$
\begin{align}
\sum_{k=0}^\infty\frac{(k+2)(k+1)}{4^k}
&=\frac2{\left(\frac34\right)^3}\\[6pt]
&=\frac{128}{27}\tag{5}
\end{align}
$$
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