How to derive the closed form of the sum of $kr^k$ $$
\sum_{k=0}^{n}kr^k = r\frac{1-(n+1)r^n + nr^{n+1}}{ (1 - r)^2 }
$$
How to derive it? I read about some finite calculus, and i understand how to tackle sums of $x^2$, $x^3$, etc.. But I don't know if the same methods can be used on this sum? 

This is an arithmetico-geometric series. 
$$\begin{align}
\sum_{k=0}^{n}kr^k=S&=0+1\cdot r+2r^2+3r^3+4r^4+\cdots+nr^n
\\rS&=\qquad \qquad \quad r^2+2r^3+3r^4+\cdots+(n-1)r^n+nr^{n+1}\\
\text{Subtracting:}\\
(1-r)S&=\qquad \quad r+\ \ r^2+\ \ r^3+\ r^4
+\cdots+\ \ r^n \qquad \quad -nr^{n+1}\\
&=\frac{r(1-r^n)}{1-r}-nr^{n+1}\\
S&=\frac {r(1-r^n)-(1-r)nr^{n+1}}{(1-r)^2}\\
&=\frac{r(1-r^n)-r(1-r)nr^n}{(1-r)^2}\\
&=r\frac{1-(n+1)r^n+nr^{n+1}}{(1-r)^2}\qquad \blacksquare\end{align}$$
