If n is a multiple of 4, the sum s=1+2i+3i2+...+(n+1)in, where i=−1, equals:<spanclass=′latex−bold′>(A)</span>1+i<spanclass=′latex−bold′>(B)</span>21(n+2)<spanclass=′latex−bold′>(C)</span>21(n+2−ni)<spanclass=′latex−bold′>(D)</span>21[(n+1)(1−i)+2]<spanclass=′latex−bold′>(E)</span>81(n2+8−4ni)