i−100+i−99+i−98+⋯+i−1+i0+i1+⋯+i99+i100
i=√−1, we start simple and try to find a pattern.
i0=1i1=ii2=−1i3=−ii4=1i5=ii6=−1i7=−i
Every 4 terms, the sequence rotates, 1,i,−1,−i
Their sum is: 1+i−1−i=0
Since every four terms cancel out, we just need to know what term the sequence starts and ends.
i−100=1i100=1
i−100+i−99+i−98+⋯+i−1+i0+i1+⋯+i99+i100=1+i−1−i+1+−⋯+1+i−1−i+1=1
I hope this helped,
Gavin