i−2−3i+4… to 100 terms =

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)

i−2−3i+4… to 100 terms =

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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$i - 2 - 3 i + 4 \dots to 100 terms =$

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$25 i$

$100 (1 - i)$

$25 (1 + i)$

$50 (1 - i)$

answer is A.

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Detailed Solution

$\begin{array}{l} S = i + 2 i^{2} + 3 i^{3} +, \dots, + 100 i^{100} \\ S \cdot i = i^{2} + 2 i^{3} +, \dots, + 99 i^{100} + 100 i^{101} \\ ∴ S (1 - i) = (i + i^{2} + i^{3} +, \dots, + 100 terms) - 100 i^{101} \\ = \frac{i (1 - i^{100})}{1 - i} - 100 i = \frac{i (1 - 1)}{1 - i} - 100 i = - 100 i \\ ∴ S = \frac{- 100 i}{1 - i} = \frac{- 100 i (1 + i)}{2} = 50 (1 - i) \end{array}$