If A=2nC0⋅2nC1+2nC12n−1C1+2nC22n−2C1+…+2nC2n−1⋅1C1 then A is
A
0
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B
2n
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C
n⋅22n
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D
1
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Solution
The correct option is Cn⋅22n A= Coefficient of x in [2nC0(1+x)2n +2nC1(1+x)2n−1+…..2nC2n−1(1+x)1] = Coefficient of x in [(1+(1+x)]2n−2nC2n = Coefficient of x in (2+x)2n = Coefficient of x in 22n(1+x2)2n=n⋅22n