Problem: If n is a multiple of 4, the sum s=1+2i+3i2+…+(n+1)in, where i=−1, equals:
Answer Choices:
A. 1+i
B. 21(n+2)
C. 21(n+2−ni)
D. 21[(n+1)(1−i)+2]
E. 81(n2+8−4ni)
Solution:
s=1+2i+3i2+…+(n+1)in
is=i+2i2+…+nin+(n+1)in+1
s(1−i)=1+i+i2+…+in−(n+1)in+1
s(1−i)=1−i1−in+1−(n+1)in+1=1−(n+1)i since in=1
∴s=1−i1−(n+1)i⋅1+i1+i=21(n+2−ni)
OR
s=1+3i2+5i4+7i6+…+(n+1)in+2i+4i3+…+nin−1
s=(1−3)+(5−7)+…((n−3)−(n−1))+n+1+i[(2−4)+(6−8)+…+(n−2−n)]
s=4n(−2)+n+1+4n(−2)i=21(n+2−ni)