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

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
