# Math Help - Root of unity sum

1. ## Root of unity sum

Let X be one of the nth roots of unity.

Find the sum: $1 + 2x + 3x^2 + ... + nx^{n-1}$

I've been trying to figure this out but nothing...
I know that sum of roots of unity is 0. But I have no clue about this one.

2. ## Re: Root of unity sum

Let $f(x)=x+x^2+...+x^n$
This is geometric progression with first member x and q=x.
Therefore, its sum is $f(x)=\frac {x(x^n-1)}{x-1}$
We obtained equality: $x+x^2+...+x^n=\frac {x(x^n-1)}{x-1}$
Differentiating with respect to x yields:
$1+2x+...+nx^{n-1}=\frac {nx^{n+1}-(n+1)x^n+1}{(x-1)^2}$
Since x is n-th root of unity then $x^n=1$ and we can write above equality as
$1+2x+...+nx^{n-1}=\frac {nx^{n+1}-(n+1)*1+1}{(x-1)^2}$
Or
$1+2x+...+nx^{n-1}=n \frac {x^{n+1}-1}{(x-1)^2}$
Again, since $x^n=1$ then $n \frac {x^{n+1}-1}{(x-1)^2}=n \frac {x x^n-1}{(x-1)^2}=n \frac {x-1}{(x-1)^2}=\frac {n}{x-1}$
Finally,
$1+2x+...+nx^{n-1}=\frac {n}{x-1}$

As can be seen from formula x=1 can't be applied to the formula.
When x=1 $1+2x+...+nx^{n-1}=1+2+3+...+n=\frac{n(n+1)}{2}$

Therefore, if x=1 $1+2x+...+nx^{n-1}=\frac{n(n+1)}{2}$, otherwise $1+2x+...+nx^{n-1}=\frac {n}{x-1}$

3. ## Re: Root of unity sum

Hello, gregx22!

$\text{Let }x\text{ be one of the }n^{th}\text{ roots of unity.}$

$\text{Find the sum: }\:S \;=\;1 + 2x + 3x^2 + \hdots + nx^{n-1}$

$\begin{array}{ccccccc}\text{We have:} & S \;=\; 1 + 2x + 3x^2 + 4x^3 + \hdots + nx^{n-1} \qquad\qquad\quad \\ \text{Multiply by }x\!: & xS \;=\; \quad\; x + 2x^2 + 3x^3 + \hdots + (n-1)x^{n-1} + nx^n \end{array}$

$\text{Subtract: }\:S - xS \;=\;\underbrace{1 + x + x^2 + x^3 + \hdots + x^{n-1}}_{\text{geometric series}} - nx^n$

$\text{The geometric series has: first term }a = 1\text{, common ratio }r = x\text{, and }n\text{ terms.}$
. . $\text{Its sum is: }\:\frac{1 - x^n}{1-x}$
$\text{But }x\text{ is an }n^{th}\text{ root of unity. }\;\text{Hence, }1 - x^n \:=\:0$

$\text{So we have: }\:S(1-x) \;=\;-nx^n \quad\Rightarrow\quad S \;=\;\frac{nx^n}{x-1} \;=\;\frac{n}{x-1}$