i−2−3i+4… to 100 terms =
50(1−i)
25i
25(1+i)
100(1−i)
S=i+2i2+3i3+,…,+100i100S⋅i=i2+2i3+,…,+99i100+100i101∴S(1−i)=i+i2+i3+,…,+100 terms −100i101=i1−i1001−i−100i=i(1−1)1−i−100i=−100i∴ S=−100i1−i=−100i(1+i)2=50(1−i)