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

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 $x:=\dfrac14.$
