The correct option is A 50(1−i)
S=i+2i2+3i3+⋯ 100i100⋯(1)
Si= i2+2i3+⋯+99i100+100i101⋯(2)
Subtracting Equation (2) from equation (1), we get
S(1−i)=i+i2+i3+⋯ i100−100i101
∵Sum of n terms of G.P Sn=a(1−rn)1−r
Here, a=i,r=i,n=100
⇒S(1−i)=i(1−i100)1−i−100i100⋅i
[∵i100=(i4)25=1]
Now, S(1−i)=0−100i
⇒S=100(−i1−i)
⇒S=100(−i(1+i)2)
∴S=50(1−i)