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
 A: Hint:
$${\frac {1}{1-x}}=\sum _{n=0}^{\infty }x^{n}\quad {\text{ for }}|x|<1\!$$
differentiate  it and multiply by $x$
$${\frac {x}{(1-x)^{2}}}=\sum _{n=0}^{\infty }nx^{n}$$
differentiate it again and multiply by $x$
$${\frac {x(x+1)}{(1-x)^{3}}}=\sum _{n=0}^{\infty }n^2x^{n}$$
A: Hint:   for $|x| \lt 1$
$$f(x) = \sum_{n=0}^{\infty} x^{n+2} = \frac{x^2}{1-x} = -1 -x + \frac{1}{1-x}$$
Now consider $f''(\frac{1}{4})$.
A: Here is another approach. Let us define the function 
$$\phi(x):=\sum_{n=0}^\infty (n+1)(n+2) x^n.$$
If we differentiate we obtain
$$\phi'(x):=\sum_{n=0}^\infty n(n+1)(n+2) x^{n-1}=\sum_{n=0}^\infty(n+1)(n+2)(n+3)x^n.$$
Now, 
$$\phi'(x)-3\phi(x)=\sum_{n=0}^\infty n(n+1)(n+2)x^{n}=\sum_{n=0}^\infty(n+1)(n+2)(n+3)x^{n+1}=x\phi'(x)$$
so
$$\phi'(x)(x-1)+3\phi(x)=0.$$
An elementary integration (with the initial condition $\phi(0)=2$) yields
$$\phi(x)=\frac2{(1-x)^3}.$$
Thus, the value you are seeking is $\phi(1/4)=128/27$.
A: 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 $x=\frac14$:
$$
\begin{align}
\sum_{k=0}^\infty\frac{(k+2)(k+1)}{4^k}
&=\frac2{\left(\frac34\right)^3}\\[6pt]
&=\frac{128}{27}\tag{5}
\end{align}
$$
A: prove with induction that $$\sum_{n=0}^k \frac{n^2+3n+2}{4^n}=\frac{1}{27} 4^{-k} \left(-9 k^2-51 k+2^{2 k+7}-74\right)$$
A: 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 $x:=\dfrac14.$
A: $$
\begin{aligned}
& S_{0} = \sum_{n=0}^{\infty} \frac{n^{0}}{4^{n}} = \sum_{n=0}^{\infty} \frac{1}{4^{n}} = \sum_{n=0}^{\infty} (1/4)^{n} = \frac{1}{1 - (1/4)} \Rightarrow \color{red}{S_{0} = \frac{4}{3}} \\ \\
& S_{1} = \sum_{n=0}^{\infty} \frac{n^{1}}{4^{n}} = \sum_{n=0}^{\infty} \frac{n + 1 - 1}{4^{n}} = \sum_{n=0}^{\infty} \frac{n + 1}{4^{n}} - \sum_{n=0}^{\infty} \frac{1}{4^{n}} = 4 \sum_{n=0}^{\infty} \frac{n + 1}{4^{n + 1}} - S_{0} \\
& \qquad = 4 \sum_{n=1}^{\infty} \frac{n}{4^{n}} - S_{0} = 4 \left[ - 0 + \sum_{n=0}^{\infty} \frac{n}{4^{n}} \right] = 4 S_{1} - S_{0} \Rightarrow \color{red}{S_{1} = \frac{1}{3} S_{0} = \frac{4}{9}} \\ \\
& S_{2} = \sum_{n=0}^{\infty} \frac{n^{2}}{4^{n}} = \sum_{n=0}^{\infty} \frac{(n + 1)^{2} - 2 n - 1}{4^{n}} = 4 \sum_{n=0}^{\infty} \frac{(n + 1)^{2}}{4^{n+1}} - 2 \sum_{n=0}^{\infty} \frac{n}{4^{n}} - \sum_{n=0}^{\infty} \frac{1}{4^{n}} \\
& \qquad = 4 S_{2} - 2 S_{1} - S_{0} = 4 S_{2} - \frac{5}{3} S_{0} \Rightarrow \color{red}{S_{2} = \frac{5}{9} S_{0} = \frac{20}{27}} \\ \\
& \sum_{n=0}^{\infty} \frac{n^{2} + 3 n + 2}{4^{n}} = S_{2} + 3 S_{1} + 2 S_{0} = \frac{20}{27} +  \frac{12}{9} +  \frac{8}{3} = \frac{20 + 36 + 72}{27} = \frac{128}{27} \\ \\
\end{aligned}
$$
