TITLE: How to derive the closed form of the sum of $kr^k$
QUESTION [6 upvotes]: $$
\sum_{k=0}^{n}kr^k = r\frac{1-(n+1)r^n + nr^{n+1}}{ (1 - r)^2 }
$$
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?

REPLY [2 votes]: Via summation by parts:
$$\sum_{k=0}^n f_k g_k = f_n G_n - \sum_{j=0}^{n-1} \left( f_{j+1}- f_j\right) G_j,\ \text{where}\ G_j = \sum_{k=0}^j g_k$$
Here we have: $f_k=k$, $g_k=r^k$, $G_j=\sum\limits_{k=0}^j r^k=\dfrac{r^{j+1}-1}{r-1}$, $f_{j+1}-f_j=\left(j+1\right)-j=1$.
Then $kr^k=nG_n-\sum\limits_{j=0}^{n-1} 1\cdot G_j$. But $$\sum\limits_{j=0}^{n-1} G_j=\sum\limits_{j=0}^{n-1}\dfrac{r^{j+1}-1}{r-1}=\dfrac{1}{r-1}\left(\sum\limits_{j=0}^{n-1}r^{j+1}-n\right)=\dfrac{1}{r-1}\left(\dfrac{r^{n+1}-r}{r-1}-n\right).$$
Finally we have $kr^k=n\cdot\dfrac{r^{n+1}-1}{r-1}-\dfrac{1}{r-1}\left(\dfrac{r^{n+1}-r}{r-1}-n\right)=\dfrac{nr^{n+2}-(n+1)r^{n+1}+r}{\left(r-1\right)^2}\blacksquare$.