Summation problems involving $k c^{-k}$ I am trying to find a better way of solving problems involving summations of the form: $\sum_{k=0}^\infty k\cdot c^{-k}$ and $\sum_{k=0}^\infty k^2 \cdot c^{-k}$  
For example, finding $\sum_{k=0}^\infty k\cdot 2^{-k}$, I used the following approach:  
$S = \frac{1}{2} + 2 \cdot \frac{1}{2^2} + 3 \cdot \frac{1}{2^3} + ....$
$S = \frac{1}{2}[1+2 \cdot \frac{1}{2} + 3 \cdot \frac{1}{2^2} + .....]$
$S = \frac{1}{2}[1+\sum_{k=0}^\infty (k+1)\cdot 2^{-k}]$
$\sum_{k=0}^\infty (k+1)\cdot 2^{-k} = \sum_{k=0}^\infty k\cdot 2^{-k} + \sum_{k=0}^\infty 2^{-k} = S + \sum_{k=0}^\infty 2^{-k}$  
Therefore, $S = 1 + \sum_{k=0}^\infty 2^{-k} = 1 + \frac{1}{1-\frac{1}{2}} = 3$  
I may have made a few calculation mistakes (I would appreciate if you could point that out), but my main concern is a simpler way to approach this kind of problems. Can I use this method to evaluate problems of the form $\sum_{k=0}^\infty k^2 \cdot c^{-k}$ ? I tried, but failed.  
I would appreciate if someone could point out the most efficient way to solve this
 A: Hint: Observe
\begin{align}
\frac{1}{1-x} = \sum^\infty_{n=0}x^n
\end{align}
when $|x|<1$. In particular, we have that
\begin{align}
\frac{1}{(1-x)^2}=\frac{d}{dx}\frac{1}{1-x} = \sum^\infty_{n=1} nx^{n-1} \ \ \Rightarrow \ \ \frac{x}{(1-x)^2} = \sum^\infty_{n=1}nx^n. 
\end{align}
A: Note that we can start with the sum (where I replaced $c$ with $x$)
$$\frac{x}{x-1}=\sum_{k=0}^\infty x^{-k}$$
Differentiating and multiplying by $x$, we get that
$$\frac{x}{(x-1)^2 }= \sum_{k=0}^{\infty} kx^{-k}$$
Doing this again
$$\frac{(x+1)x}{(x-1)^3 }= \sum_{k=0}^{\infty} k^2x^{-k}$$
If you continue this process your denominator will not change, and your numerator will be an $n$th degree polynomial, where $n$ is the power of $k$ on the RHS
A: The short version:
$$S_2:=\sum_{k=0}^\infty k^2a^{-k}=\sum_{k=1}^\infty(k-1)^2a^{1-k}=a\left(\sum_{k=0}^\infty(k-1)^2a^{-k}-1\right)=a(S_2-2S_1+S_0-1)$$
and
$$S_2=a\frac{2S_1-S_0+1}{a-1}.$$

The full version:
We can solve this for any power, using
$$S_d:=\sum_{k=0}^\infty k^da^{-k}=\sum_{k=1}^\infty(k-1)^da^{1-k}=a\left(\sum_{k=0}^\infty(k-1)^da^{-k}-(-1)^d\right).$$
Then by the Binomial theorem,
$$(k-1)^d=\binom d0k^{d}-\binom d1k^{d-1}+\binom d2k^{d-2}\cdots+\binom dd(-1)^d,$$
and by subtraction we can cancel the powers $k^d$, giving
$$\left(1-\frac1a\right)S_d=\binom d1S_{d-1}-\binom d2S_{d-2}\cdots+\binom dd(-1)^{d-1}S_0+(-1)^d.$$
In particular
$$\left(1-\frac1a\right)S_0=1,\\
\left(1-\frac1a\right)S_1=S_0-1,\\
\left(1-\frac1a\right)S_2=2S_1-S_0+1,\\
\left(1-\frac1a\right)S_3=3S_2-3S_1+S_0-1,\\
\left(1-\frac1a\right)S_4=4S_3-6S_2+4S_1-S_0+1,\\
\cdots$$
With $a=2$, $$S_0=2,S_1=2,S_2=6,S_3=26,S_4=150,\cdots$$
