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Question

The coefficient of xn in the polynomial
(x+2n+1C0)(x+2n+1C1)(x+2n+1C2)+....(x+2n+1Cn)

A
2n+1
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B
22n+11
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C
22n
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D
22n+1+1
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Solution

The correct option is C 22n
The required coefficient will be
2n+1C0+2n+1C1+2n+1C2+2n+1C3...2n+1Cn
Now
2n+1Cr=2n+1C2n+1r
Now
(1+x)2n+1|x=1=2n+1C0+2n+1C1+2n+1C2+2n+1C3...2n+1C2n
The middle terms will 2n+1Cn and 2n+1Cn+1
Hence
22n+1=2n+1C0+2n+1C1+2n+1C2+2n+1C3...2n+1C2n
=2[2n+1C0+2n+1C1+2n+1C2+2n+1C3...2n+1Cn1]+2n+1Cn+2n+1Cn+1
=2[2n+1C0+2n+1C1+2n+1C2+2n+1C3...2n+1Cn]
Hence
22n+1=2[2n+1C0+2n+1C1+2n+1C2+2n+1C3...2n+1Cn]
2n+1C0+2n+1C1+2n+1C2+2n+1C3...2n+1Cn=22n
Thus the coefficient is 2n.

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