Let $a\ne1$ be an nth root of identity, show $1+2a+3a^2+\dots + na^{n-1} = \frac{n}{a-1}$.

Question

Let $a\ne1$ be an nth root of identity, show $1+2a+3a^2+\dots + na^{n-1} = \frac{n}{a-1}$.

I took $S= 1+2a+3a^2+\dots+na^{n-1}$.

I have a hint that says I should compute $(1-a)S$ and then use a geometric series. I can see we have a geometric series with $a$, but I do not understand why I would compute $(1-a)S$ or how this problem works out. Or what is the point of this problem (what I am supposed to take away from it).

EDIT: I also do not understand why I would get down-voted for what I believe to be a legitimate question. It's not like I am asking for an answer to a homework question. I am simply trying to understand what I am doing and perhaps gain some motivation for these concepts.

For example, I did attempt to compute $(1-a)S$, where $S = \frac{1-a^{n}}{1-a}$ (because $S$ is a finite geometric sum), but if I cancel the $(1-a)$ terms I am left with $n(1-a^{n}) \ne \frac{n}{a-1}$.

Answer 1

Hint: It sounds like you may have made an algebraic slip when computing $(1-a)S$ (don't worry, it happens to all of us every now and then!). Try computing it again and you should get a geometric series plus a negative term. Remember that $a^n = 1$ and try substituting that into what you have. It should fall out pretty easily. Hope this helps!

Answer 2

Another way:

$1+x+x^2+\cdots+x^n=\frac{x^{n+1}-1}{x-1}$. Take derivative $1+2x+\cdots+nx^{n-1}=\frac{(n+1)x^{n}(x-1)-(x^{n+1}-1)}{(x-1)^2}$. Since $a$ is $n$-th root of unity. We get $\frac{n}{a-1}$.