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Can anybody help me to show that when $x\neq 1$

$$\large \sum_{i=0}^{n-1} i\, x^i = \frac{1-n\, x^{n-1}+(n-1)\,x^n}{(1-x)^2}$$

Did
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Aharon
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    Your equation is off on the right-hand side by a multiple of x. – SWilliams Apr 21 '15 at 03:06
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    Why not prove it by induction? – GEdgar Apr 21 '15 at 03:10
  • because it is gonna be complicated if you use induction – Aharon Apr 21 '15 at 03:11
  • See: http://math.stackexchange.com/questions/180198/what-is-the-sum-of-sum-limits-i-1nipi, http://math.stackexchange.com/questions/87030/proving-sum-limits-i-0n-i-2i-1-n1-2n-1-by-induction, http://math.stackexchange.com/questions/11464/how-to-compute-the-formula-sum-r-1d-r-cdot-2r, http://math.stackexchange.com/questions/90637/what-is-the-limit-of-sum-limits-n-1-inftyn2-3nsequences-and-series, http://math.stackexchange.com/questions/405332/how-to-calculate-sum-n-1-infty-n-an – Martin Sleziak Apr 21 '15 at 04:58
  • You can find several answers giving proof by induction, for example this one. – Martin Sleziak Apr 21 '15 at 05:03
  • Please do not deface your question, even if it is currently marked as duplicate. – Did Apr 26 '15 at 06:53

3 Answers3

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We have $$\sum_{i=0}^{n-1} x^i = \dfrac{1-x^n}{1-x}$$ Differentiate both sides to obtain your answer.

Adhvaitha
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  • can you explain more by showing the details, since it is not clear for me? – Aharon Apr 21 '15 at 02:24
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    Since $\dfrac{\mathrm d}{\mathrm d x}\sum\limits_{i=0}^{n-1} x^i = \sum\limits_{i=0}^{n-1} i x^{i-1}$ , therefore $\sum\limits_{i=0}^{n-1} i x^i = x\cdot\dfrac{\mathrm d}{\mathrm d x}\left(\dfrac{1-x^n}{1-x}\right)$ – Graham Kemp Apr 21 '15 at 02:32
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The following chain of identities gives the closed form without using calculus.

$$\sum_{i=1}^n i x^i= \sum_{k=0}^{n-1} x^{n-k} \sum_{i=0}^k x^i=$$ $$=\sum_{k=0}^{n-1} x^{n-k} \frac{x^{k+1}-1}{x-1}=\frac{1}{x-1}\sum_{k=0}^{n-1}(x^{n+1}-x^{n-k})=$$ $$=\frac{x^n}{x-1}\sum_{k=0}^{n-1}(x-x^{-k})=\frac{x^n}{x-1}(n x-\frac{\frac{1}{x^n}-1}{\frac{1}{x}-1})=$$ $$=\frac{1}{x-1}(n x^{n+1}-\frac{x-x^{n+1}}{1-x})=\frac{nx^{n+2}-(n+1)x^{n+1}+x}{(x-1)^2}$$

In response to comments: I have indexed the sum in a slightly different manner than in the question asked. To derive the formula asked for, just notice that $$\sum_{i=0}^{n-1} i x^i=\sum_{i=1}^n i x^i -n x^n=\frac{nx^{n+2}-(n+1)x^{n+1}+x}{(x-1)^2}-nx^n=\frac{nx^{n+2}-(n+1)x^{n+1}+x-n x^n(x-1)^2}{(x-1)^2}=\frac{nx^{n+2}-(n+1)x^{n+1}+x-nx^{n+2}+2nx^{n+1}-nx^n}{(x-1)^2}=\frac{(n-1)x^{n+1}-nx^{n}+x}{(x-1)^2}$$

Henrik
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  • This answer is not in the form $\frac{1-n, x^{n-1}+(n-1),x^n}{(1-x)^2}$, which is what OP was asking for. Please revise – lucidgold Apr 22 '15 at 02:35
  • @lucidgold I find it highly unlikely that the OP would not be able to derive his formula by using the formula derived in this answer. But nevertheless I have revised the answer and indicated how to do that. – Henrik Apr 22 '15 at 13:28
  • I like this answer. It's the most natural way for dealing with summation of an (a+kn) sequence times a cq^n sequence. – Vim Apr 26 '15 at 07:05
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$\sum\limits_{i=0}^{n-1}ix^i = \sum\limits_{i=0}^{n-1}x\dfrac{d}{dx}x^i = x\dfrac{d}{dx}\sum\limits_{i=0}^{n-1}x^i = x\dfrac{d}{dx} \left( \dfrac{1-x^n}{1-x} \right) $ as Graham points out.

But $x\dfrac{d}{dx} \left( \dfrac{1-x^n}{1-x} \right)= x \ \dfrac{1 - nx^{n-1} + (n-1)x^n}{(1-x)^2} = \dfrac{x - nx^n + (n-1)x^{n+1}}{(1-x)^2}$.

SWilliams
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