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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

Thanks!

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

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  • $\begingroup$ 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 at 23:25
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Hint:

  • The Geometric series...

  • Differentiation.

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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.

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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 $x$.

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  • 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

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