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Let $w≠1$ be an $n$-th root of unity, i.e., $w^n-1=0$. Show that $1+2w+3w^2+\dots+nw^{n-1}=-\frac n{1-w}$.

My question is how to relate $w, w^2, \dots,w^{n-1}$ with $w^n$?

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up vote 5 down vote accepted

Hint: Let $x=1+2w+3w^2+\cdots +nw^{n-1}$. Then $$wx=w+2w^2+3w^3+\cdots +(n-1)w^{n-1}+nw^n.$$

Consider $x-wx$.

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Then it remains to show, and how to show 1+w+w^2+...+w^(n-1)=0? – i_a_n Jan 9 '13 at 2:30
$(1+w+\cdots+w^{n-1})(1-w)=1-w^n$. – André Nicolas Jan 9 '13 at 3:10

$$1+2x+3x^2+\cdots+nx^{n-1}=\frac{x+x^2+x^3+\cdots+x^n}{dx}$$ $$=\frac{d\frac{x(x^n-1)}{x-1}}{dx}$$ $$=x(x^n-1)\left(\frac{-1}{(x-1)^2}\right)+\frac1{x-1}\{(n+1)x^n-1)\}$$

If $x\ne1, x^n=1$ $$1+2x+3x^2+\cdots+nx^{n-1}=\frac n{x-1}$$

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