 Finite weighted geometric sum

Differentiate the finite geometric series ${\sum }_{n=0}^N{r}^n=(1-r^{N+1})/(1-r)$ with respect to $r$, then multiply by $r$. This proves the title identity for $r\ne 1$ without any infinite-series differentiation. At $r=1$, the sum is $N(N+1)/2$. For $|r|<1$, both $N{r}^N$ and $N{r}^{N+1}$ tend to zero, yielding ${\sum }_{n=1}^{\infty }n{r}^n=r/(1-r{)}^2$.