If $w$ be nth root of unity , then $1+2w+3w^2+\dots+nw^{n-1}$ is equal to? I tried it by letting expression $1+2w+3w^2+\dots+nw^{n-1}= x$ and then multiplying $w$ both sides . I subtracted equation 1 from 2 but it does not seems to help me because i have just started learning this topic and i am weak in properties . Can anyone help me out here ?
 A: Recall that $$\frac{z^n-1}{z-1} = 1+z+\cdots+z^{n-1}, \quad z \ne 1$$ so upon multiplying both sides by $z$ and differentiating, we obtain $$\frac{d}{dz}\left[\frac{z(z^n-1)}{z-1}\right] = 1+2z+3z^2+\cdots+nz^{n-1}.$$  After computing the derivative, evaluate the LHS for a primitive root of unity $z = e^{2\pi i/n}$.

Alternatively, by Abel summation:  let $$f(z) = 1+2z+3z^2+\cdots+nz^{n-1},$$ so that $$\begin{align*} zf(z) - f(z) &= (z+2z^2+3z^3+\cdots+nz^n) - (1+2z+3z^2+\cdots+nz^{n-1}) \\ &= nz^n - (1+z+z^2+\cdots+z^{n-1}) \\ &= nz^n - \frac{z^n-1}{z-1}. \end{align*}$$  It immediately follows that $$f(z) = \frac{nz^n(z-1) - (z^n-1)}{(z-1)^2},$$ and the same substitution as in the above method easily results in $$f(w) = \frac{n(w-1)}{(w-1)^2} = \frac{n}{w-1}.$$
A: $$\sum_{k=0}^n w^k = \frac{w^{n+1}-1}{w-1}\implies \sum_{k=0}^nkw^{k-1} = 
\frac{(n+1)(w-1)w^n - \left(w^{n+1} - 1\right)}{\left(w-1\right)^2}$$
We know that $w^n = 1$, and so we can reduce this to be:
$$\frac{(n+1)(w-1) - \left(w - 1\right)}{\left(w-1\right)^2} = \frac n{w - 1}$$
A: Applying twice your method of multiplying by $w$ and subtracting, and using $w^n=1$ 
$$1+2w+3w^2+\dots+nw^{n-1}= x$$ 
$$w+2w^2+3w^3+\dots+nw^{n}= xw$$
$$1+w+w^2+\dots w^{n-1}= x(1-w)+n$$
$$w+w^2+w^3+\dots+ w^{n}= x(1-w)w+nw$$
$$1-1=x(1-w)w+nw$$
$$x=\frac{-n}{1-w}$$
unless $w=1$ (prohibiting the division), in which case the sum is $n(n+1)/2$; you can ignore the possibility that $w=0$.
