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Question

If (1+x)n=C0+C1x+C2x2+...+Cnxn, then C0C2+C1C3+C2C4+...+Cn−2Cn=

A
(2n)!(n!)2
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B
(2n)!(n1)!(n+1)!
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C
(2n)!(n2)!(n+2)!
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D
none of these
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Solution

The correct option is C (2n)!(n2)!(n+2)!
Consider the following series.
mC0nCr+mC1nCr1+...mCrnC0
=Coefficient of xr in (1+x)m.(x+1)n
=coefficient of xr in (1+x)m+n
=m+nCr
In the above case, r=n2 and m=n
Hence
2nCn2
=2n!(2n(n2))!(n2)!
=2n!(n+2)!(n2)!

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