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? 

You can prove the relation using induction:
Basis (n=0): clear.
Inductive step: 
$$
r\frac{1 - (n+1)r^{n} + nr^{n+1}}{(1-r)^2} + (n+1)r^{n+1} = r\frac{1-(n+1)r^{n} + nr^{n+1} + (n+1)r^{n} - 2(n+1)r^{n+1} + (n+1)r^{n+2} }{(1-r)^2} = r\frac{1-(n+2)r^{n+1} + (n+1)r^{n+2}}{(1-r)^2}
$$
Then, it follows. Another way to come up with it is differentiating the regular formula for the geometric progression:
$$
\frac{d}{dr}\sum_{k=0}^{n} r^{k} = \sum_{k=0}^{n} kr^{k-1} = \frac{d}{dr}\frac{1-r^{n+1}}{1-r} = \frac{-(n+1)r^{n}(1-r) + 1 - r^{n+1}}{(1-r)^2} = \frac{1-(n+1)r^{n} + nr^{n+1}}{(1-r)^2}
$$
Then, just multiply the whole thing by $r$ and you're done.
