Date: Jan 30, 2007 8:35 PM
Author: Michael Olea
Subject: Re: complex numbers
jennifer wrote:

> How would i evaluate the following:

>

> 1+2w+3w^2+...+nw^n-1, where w is an nth root of unity.

>

> apparently, the solution is: n(n+1)/2 if w= 1

>

> I don't know how they got this? Any hints please?

I can give you a hint for this part. When w=1 the sum is:

1+2+3+...+n

One hint is to notice the pattern:

1+n = n+1

2+(n-1) = n+1

3+(n-2)= n+1

...

And treat the cases n is odd and n is even separately.

Another way to look at it is to consider an nxn (square) matrix with 1's on

the main diagonal and upper triangle, and 0's on the lower triangle, and

figure out a formula for the number of 1's. For exmple, if n=5:

11111

01111

00111

00011

00001

-- Michael