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