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This question already has an answer here:

I would like to compute the following sum : $$\sum_{i=1}^k{i\cdot c^i}$$ where $k$ and $c$ are any real numbers.

I know how to compute $i$ and $c^i$ separately but I don't know how to do it when they are multiplied together.

Thanks a lot for the help!

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marked as duplicate by Hans Lundmark, Community Jan 27 at 21:06

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.

  • $\begingroup$ Welcome to Maths SX! I suppose you mean you want to compute this sum? $\endgroup$ – Bernard Jan 27 at 18:23
  • $\begingroup$ Oh yes thank you ! $\endgroup$ – Louis-Simon Cyr Jan 27 at 19:31
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Hint:

Do some analysis: $$\sum_{i=1}^k ic^i=c\sum_{i=1}^kic^{i-1}=c\Bigl(\sum_{i=0}^k c^i\Bigr)'. $$

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Let $Z=\sum_{i=1}^kc^i$, then (derivative with respect to $c$), $Z'=\sum_{i=1}^kic^{i-1}$. When multiplied with $c$, we get your question, i.e. we want $cZ'$. You can easily substitute $Z=\frac{1-c^{k+1}}{1-c}-1$.

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Hint: Let $a,q$ be real numbers with $q\ne 1$: $$a + aq + aq^2 + \ldots +aq^{n-1} = a\frac{q^n-1}{q-1}.$$

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    $\begingroup$ This is a true fact, but it is irrelevant to the question. OP is trying to compute the sum $$1c^1 + 2 c^2 + 3 c^3 + \cdots + k c^k$$ The coefficients on the powers of $c$ are not constant. $\endgroup$ – Blue Jan 27 at 20:15

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