∑[ i^n, n, 1, 99]=-1, because[i^1 + i^2 +i^3 + i^4]=0 + [i^5+i^6+i^7+i^8]=0+0=0. This pattern continues until i^96, which is a multiple of 4. i^97=i, i^98=-1, i^99=-i. So add the last 3 together:
i + (-1) + (-i) = -1 - the two "i" cancel leaving -1 as the answer.
