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i23i+4 to 100 terms =

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a
50(1−i)
b
25i
c
25(1+i)
d
100(1−i)

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detailed solution

Correct option is A

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)


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