Formula for calculating $\sum_{n=0}^{m}nr^n$

I see that one of the tags is pre-calculus, so here is a way to answer the question that does not use differentiation:

$S = r + 2r^2 +3r^3 +\dots + (m-1)r^{m-1}+mr^m $
$rS = \ \ \ \ r^2 +2r^3 +\dots + (m-2)r^{m-1}+(m-1)r^m + mr^{m+1} $

Subtracting the bottom line from the top, we get
$$(1-r)S = r+r^2 +r^3 + \dots + r^{m-1} + r^m -mr^{m+1} .$$ But using the formula for the sum of a geometric series, we have that $$(1-r)S = \frac{r(1-r^m)}{1-r} -mr^{m+1}.$$ Dividing by $(1-r)$, we have $$ S=\frac{r(1-r^m)}{(1-r)^2} -\frac{mr^{m+1}}{1-r}. $$ (Obviously, for this to hold, one needs $r \neq 1$. If $r=1$, then we are looking at $\sum_{n=1}^m n= \frac{m(m+1)}{2}$.)

Observe that your formula $\sum_{n=0}^{m}nr^n$ can be obtained from $\sum_{n=0}^{m}x^n$ by applying $x\frac d{dx}$ (deriving and then multiplying by $x$) and then substituting $r$ for $x$. Now for geometric series one has the well known formula $$ \sum_{n=0}^mx^n=\frac{x^0-x^{m+1}}{1-x} $$ and applying $x\frac d{dx}$ to the right hand side gives $$ \frac{x-(m+1)x^{m+1}+mx^{m+2}}{(1-x)^2} = x\frac{1-x^m+mx^m(x-1)}{(1-x)^2} =x\left(\frac{1-x^m}{(1-x)^2}-\frac{mx^m}{1-x}\right) $$ so that your answer should be $$ \sum_{n=0}^{m}nr^n=r\left(\frac{1-r^m}{(1-r)^2}-\frac{mr^m}{1-r}\right). $$ For $r=2$ this gives $2(1+(m-1)2^m)$, in accordance with what you found.

Hint: