<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] One may start with the standard finite evaluation:
$$
1+x+x^2+...+x^n=\frac{1-x^{n+1}}{1-x}, \quad |x|<1. \tag1
$$ Then by differentiating $(1)$ we have
$$
1+2x+3x^2+...+nx^{n-1}=\frac{1-x^{n+1}}{(1-x)^2}+\frac{-(n+1)x^{n}}{1-x}, \quad |x|<1, \tag2
$$ by differentiating once more one gets 
$$
2\times 1+3\times 2 x^2+...+n\times (n-1)x^{n-2}=\frac{2-x^{n-1}\left(n+n^2 (1-x)^2+2x-nx^2\right)}{(1-x)^3},\tag3
$$ then by making $n \to +\infty$ in $(3)$, using $|x|<1$, one obtains 

$$
\sum_{n=0}^\infty(n+1)\cdot n \cdot x^n=\frac{2 x}{(1-x)^3},\quad |x|<1, \tag4
$$ 

from which one gets the desired series by putting $x:=\dfrac14.$
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