Compute \(i^600+i^599+....+i+1 = 1.\) Where \(i^2=-1\).
\(\text{I assume this is supposed to be}\\ \sum \limits_{k=0}^{600}i^k\)
The sum of any 4 adjacent terms of this series is 0.
600 is divisible by 4.
There are 601 terms, 600 of which will sum to zero leaving the last (or first) term i600 = i0 = 1
