Question

# The coefficient of x50 in the expansion of (1+x)1000+2x(1+x)999+3x2(1+x)998+......+1001 x1000

A
1000C50
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B
1001C50
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C
1002C50
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D
21001
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Solution

## The correct option is C 1002C50Let, S=(1+x)1000+2x(1+x)999+3x2(1+x)998+......+1001 x1000 x1+xS=x(1+x)999+2x2(1+x)998+........+1000x1000+1001x10011+x Substract above equations. (1−x1+x)S=(1+x)1000+(1+x)999+x2(1+x)998+....+x1000−1001x10011+x ⇒S=(1+x)1001+x(1+x)1000+x2(1+x)999+......+x1000(1+x)−1001x1001 =(1+x)1001[(x1+x)1001−1]x1+x−1−1001 x1001 [Sum of G.P] =(1+x)1002−x1002−1002 x1001 ∴ coefficient of x50 in S= coefficient of x50 in [(1+x)1002−x1002−1002x1001]=1002C50

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