The sequence S=i+2i2+3i3+… upto100 terms simplifies to where i=−1

s=i+2i2+3i3+….+99i99+100i100 is=i2+2i3+….+98i99+99i100+100i101 S(1−i)=i+i2+i3+…..+i100−100,i101 (1−i)S=−100i1−i(1+i)(1+i)⇒s=50(1−i)

The sequence S=i+2i2+3i3+… upto100 terms simplifies to where i=−1

s=i+2i2+3i3+….+99i99+100i100 is=i2+2i3+….+98i99+99i100+100i101 S(1−i)=i+i2+i3+…..+i100−100,i101 (1−i)S=−100i1−i(1+i)(1+i)⇒s=50(1−i)

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The sequence $S = i + 2 i^{2} + 3 i^{3} + \dots$ upto100 terms simplifies to where $i = \sqrt{- 1}$

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$50 (i - 1)$

$100 (1 + i)$

$100 (1 - i)$

$50 (1 - i)$

answer is A.

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Detailed Solution

$s = i + 2 i^{2} + 3 i^{3} + \dots . + 99 i^{99} + 100 i^{100} i s = i^{2} + 2 i^{3} + \dots . + 98 i^{99} + 99 i^{100} + 100 i^{101} S (1 - i) = i + i^{2} + i^{3} + \dots .. + i^{100} - 100, i^{101} (1 - i) S = \frac{- 100 i}{1 - i} \frac{(1 + i)}{(1 + i)} \Rightarrow s = 50 (1 - i)$