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Re: complex numbers
Posted:
Jan 30, 2007 9:45 PM
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Michael Olea wrote:
> 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.
Hm. I have, up till now, always had to stop and think a second or two about
the boundary conditions for n odd or even, but it occurs to me there is a
simpler way to see the result. Line up the numbers like this (monospaced
font recommended):
1 2 3 ... (n-2) (n-1) n
n (n-1) (n-2) ... 3 2 1
==============================
Add the top row to the bottom row, which gives n summands all adding up to
(n+1). That will be twice the value of the desired result...
-- Michael
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