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Question

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


A
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B
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C
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D
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Solution

The correct option is C
Let, 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.
(1x1+x)S=(1+x)1000+(1+x)999+x2(1+x)998+....+x10001001x10011+x
S=(1+x)1001+x(1+x)1000+x2(1+x)999+......+x1000(1+x)1001x1001
=(1+x)1001[(x1+x)10011]x1+x11001 x1001
[Sum of G.P]
=(1+x)1002x10021002 x1001
coefficient of x50 in S= coefficient of x50 in [(1+x)1002x10021002x1001]=1002C50

Mathematics

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