> 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...