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

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}
