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This question already has an answer here:

Is there a formula to get $r+2r^2+3r^3+\dots+nr^n$ provided that $|r|<1$? This seems like the geometric "sum" $r+r^2+\dots+r^n$ so I guess that we have to use some kind of trick to get it, but I cannot think of a single one. Can you please help me with this problem?

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marked as duplicate by user147263, Rolf Hoyer, Prasun Biswas, Yiorgos S. Smyrlis calculus May 6 '15 at 4:25

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ If you take it to infinity then yes. $\endgroup$ – Jorge Fernández Hidalgo May 6 '15 at 0:02
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    $\begingroup$ This is surely a duplicate. $\endgroup$ – Akiva Weinberger May 6 '15 at 0:06
  • $\begingroup$ @Gamamal There is a well-known formula for the finite sum, too. $\endgroup$ – Akiva Weinberger May 6 '15 at 0:06
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    $\begingroup$ You know the sum of a finite geometric progression. Differentiate and multiply by $r$ $\endgroup$ – André Nicolas May 6 '15 at 0:07
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    $\begingroup$ Another way: Knowing that $(n+2)-2(n+1)+(n)=0$, calculate $S-2S+S$ in a clever way. $\endgroup$ – Akiva Weinberger May 6 '15 at 0:09
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We have $$1+r+r^2 + \cdots + r^n = \dfrac{r^{n+1}-1}{r-1}$$ Differentiating this once, we obtain $$1+2r + 3r^2 + \cdots + nr^{n-1}= \dfrac{nr^{n+1}-(n+1)r^n + 1}{(r-1)^2}$$ Multiply the above by $r$ to obtain what you want.

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    $\begingroup$ Note that this works even if $|r| \ge 1$ $\endgroup$ – GFauxPas May 6 '15 at 0:43
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HINT: $r+2r^2+3r^3+... +nr^n=(r+r^2+\dots+r^n)+(r^2+r^3+\dots+r^n)+\dots+(r^n)$ and compute values in parentheses.

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  • $\begingroup$ Oh dayum, this is nice. $\endgroup$ – Jorge Fernández Hidalgo May 6 '15 at 0:09
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    $\begingroup$ @Gamamal Thank you! Without differentiation it is possible to understand on a lower level of mathematical education. $\endgroup$ – Przemysław Scherwentke May 6 '15 at 0:12
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You can use the following approach: $$ \sum_{k=1}^n k r^k = r\sum_{k=1}^n k r^{k-1} = r\left(\sum_{k=1}^n r^{k}\right)^\prime = r\left(r\frac{1-r^n}{1-r}\right)^\prime $$

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What I am thinking of is this: $$r+2r^2+3r^3+...+nr^n = \sum_{k=1}^{n}r^k + \sum_{k=2}^nr^k + ... \sum_{k=n}^nr^k$$ Each is a geometric series, so the sum would be $$\frac{1-r^n}{1-r} + r\frac{1-r^{n-1}}{1-r} + ... + r^n$$.

Hope this helps.

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\begin{align*}r+2r^2+3r^3+\ldots+nr^n&=r(1+2r+3r^2+\ldots+nr^{n-1})=r(1+r+r^2+\dots+r^n)'\\&=r\biggl(\frac{1-r^{n+1}}{1-r}\biggl)'=r\,\frac{nr^{n+1}-(n+1)r^n+1}{(1-r)^2}. \end{align*}

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