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The sum of the series $1 + 2 \cdot 2 + 3 \cdot 2^{2} + 4 \cdot 2^{3} + 5 \cdot 2^{4} + \dots + 100 \cdot 2^{99} is$

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a

99⋅2100-1

b

100⋅2100

c

99⋅2100

d

99⋅2100+1

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detailed solution

Correct option is D

Let S=1+2⋅2+3⋅22+4⋅23+5⋅24 +…+100⋅299∴ 2S=1⋅2+2⋅22+3⋅23+…+99⋅299 +100⋅2100 Substracting, we get −S=1+1⋅2+1⋅22+…+1⋅299−100⋅2100=1+2+22+…+299−100⋅2100=12100−12−1−100⋅2100=2100−1−100⋅2100∴S=100⋅2100−2100+1=99⋅2100+1.

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