I want to know the general formula for $\sum_{n=0}^{m}nr^n$ for some constant r and how it is derived.

For example, when r = 2, the formula is given by: $\sum_{n=0}^{m}n2^n = 2(m2^m - 2^m +1)$ according to http://www.wolframalpha.com/input/?i=partial+sum+of+n+2%5En


  • $\begingroup$ Certainly asked here several times. Can't find a link right now. $\endgroup$ – lhf Mar 13 '12 at 11:22
  • $\begingroup$ Brilliant answers from everyone. thank you all so much $\endgroup$ – hollow7 Mar 16 '12 at 13:08
  • 1
    $\begingroup$ @dragoncharmer Either you tag this [algebra-precalculus] or you tag it [taylor-expansion]. Those are quite mutually exclusive tags. $\endgroup$ – Pedro Tamaroff May 2 '12 at 1:46
  • $\begingroup$ See also: What is the sum of $\sum\limits_{i=1}^{n}ip^i$? $\endgroup$ – Martin Sleziak Sep 23 '17 at 6:12

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 $\sum_{n=1}^m n= \frac{m(m+1)}{2}$.)

  • $\begingroup$ You do not even need to know the formula for a geometric series. From your series expression for $(1-r)S$ you could repeat your multiply and subtract steps to get $(1-r)^2S = r -(m+1)r^{m+1}+mr^{m+2}$ and thus your final result $\endgroup$ – Henry Mar 18 '19 at 23:25

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.



  • The Geometric series...

  • Differentiation.


I suppose you are familiar with the sum of an geometric progression: $$ 1+x+x^2+\dots+x^m=\frac{x^{m+1}-1}{x-1}. $$ Take derivatives an multiply by $x$.

  • 3
    $\begingroup$ Typo: sum of Geometric progression $\endgroup$ – Learner Mar 14 '12 at 16:07
  • $\begingroup$ @Learner I just edited $\endgroup$ – Jingjie YANG Oct 24 '18 at 19:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.