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Question

The value of C0+3C1+5C2+7C3+.............+(2n+1)Cn is equal to

A
2n
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B
2n+n.2n1
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C
2n.(n+1)
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D
none of these
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Solution

The correct option is C 2n.(n+1)
Consider the following
(1+x2)n=1+nC1x2+nC2x4+nC3x6...nCnx2n
Multiplying both sides with x, we get
x(1+x2)n=x+nC1x3+nC2x5+nC3x7...nCnx2n+1
Differentiating both sides with respect to x, we get
[(1+x2)n+nx(2x)(1+x2)n1]x=1=1+3nC1+5nC2+...(2n+1)nCn
Hence
1+3nC1+5nC2+...(2n+1)nCn
=[(1+x2)n+nx(2x)(1+x2)n1]x=1
=2n+2n(2n1)
=2n+n2n
=2n(n+1)

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