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Calculate $1+2\epsilon+3\epsilon^{2}+\cdots+n\epsilon^{n-1}$ Where $\epsilon$ is nth root of unity. There is a hint that says: multiply by $(1-\epsilon)$

Doing this multiplication I get: $1+\epsilon+\epsilon^{2}+\epsilon^{3}+\cdots+\epsilon^{n-1}-n$ The answer is: $-\frac{n}{1-\epsilon} $if $ \epsilon\neq1$ and $\frac{n(n+1)}{2}$ if $\epsilon=1$. I can't see how to get this result...

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    A formula for $\sum\limits_{r = 1}^{d} r x^{r}$ si given in [this answer](http://math.stackexchange.com/questions/11464/how-to-compute-the-formula-sum-limits-r-1d-r-cdot-2r/11549#11549). (It is more complicated, since it is a formula for general $x$; without using $x^d=1$).2012-10-03

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If $\epsilon=1$, $1+2\epsilon+3\epsilon^{2}+\cdots+n\epsilon^{n-1}=\sum_{k=1}^nk=\frac{n(n+1)}2$ by the familiar formula for the sum of the first $n$ positive integers (or more generally by the formula for the sum of a finite arithmetic series).

If $\epsilon\ne 1$, you’ve shown that

$(1-\epsilon)(1+2\epsilon+3\epsilon^{2}+\cdots+n\epsilon^{n-1})=\sum_{k=0}^{n-1}\epsilon^k-n=\frac{1-\epsilon^n}{1-\epsilon}-n=-n\;,$

where I’ve used the formula for the sum of a finite geometric series and the fact that $\epsilon^n=1$. It follows immediately that

$1+2\epsilon+3\epsilon^{2}+\cdots+n\epsilon^{n-1}=-\frac{n}{1-\epsilon}$ in this case.

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    @Mykolas: You’re welcome.2012-10-03
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Of course, when $\epsilon=1$, $ \begin{align} &1+2\epsilon+3\epsilon^2+4\epsilon^3+\dots+n\epsilon^{n-1}\\ &=1+2+2+4+\dots+n\\ &=\frac{n(n+1)}{2} \end{align} $ When $\epsilon\ne1$ $ \left. \begin{align} 1+\epsilon+\epsilon^2+\epsilon^3+\dots+\epsilon^{n-1}&=\frac{1-\epsilon^n}{1-\epsilon}\\ \epsilon+\epsilon^2+\epsilon^3+\dots+\epsilon^{n-1}&=\frac{\epsilon-\epsilon^n}{1-\epsilon}\\ \epsilon^2+\epsilon^3+\dots+\epsilon^{n-1}&=\frac{\epsilon^2-\epsilon^n}{1-\epsilon}\\ \epsilon^3+\dots+\epsilon^{n-1}&=\frac{\epsilon^3-\epsilon^n}{1-\epsilon}\\ &\vdots\\ \epsilon^{n-1}&=\frac{\epsilon^{n-1}-\epsilon^n}{1-\epsilon}\\ \end{align} \right\}\mbox{ $n$ equations} $ Summing these $n$ equations yields $ \begin{align} 1+2\epsilon+3\epsilon^2+4\epsilon^3+\dots+n\epsilon^{n-1} &=\frac{\frac{1-\epsilon^n}{1-\epsilon}-n\epsilon^n}{1-\epsilon}\\ &=\frac{n}{\epsilon-1} \end{align} $

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$1+2\epsilon+3\epsilon^{2}+\cdots+n\epsilon^{n-1} = f'(\epsilon)$, where $f(x)=1+x+x^2+\cdots+x^n= \dfrac{x^{n+1}-1}{x-1}$.

Now $f'(x)=\dfrac{(n+1)x^n(x-1)-(x^{n+1}-1)}{(x-1)^2} $ and so $f'(\epsilon)=\dfrac{(n+1)(\epsilon-1)-(\epsilon-1)}{(\epsilon-1)^2}=\dfrac{n}{\epsilon-1}$, if $\epsilon\neq 1$.

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$1+\epsilon+\epsilon^{2}+\epsilon^{3}+\cdots+\epsilon^{n-1}$ is a geometric series and its summation can be computed as $\frac{1-\epsilon^n}{1-\epsilon}$ for $\epsilon\neq 1$.