Differentiation to Solve Modified Sum of Binomial Coefficients
The sum of th...
Question
The sum of the series 1+2⋅2+3⋅22+4⋅23+5⋅24+⋯+100⋅299, is
A
99⋅2100
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B
99⋅2100+1
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C
100⋅2100
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D
none of these
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Solution
The correct option is B99⋅2100+1 Let S=1+2⋅2+3⋅22+4⋅23+5⋅24+⋯100⋅299 .................. (1) Multiply (1) by 2 2S=2+2⋅22+3⋅23+⋯100⋅2100............(2) Subtracting (1) and (2) gives (1−2)S=1+2+22+23+⋯299−100⋅2100 ⇒−S=2100−12−1−100.2100⇒−S=−1−99.2100 ⇒S=1+99⋅2100