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Question

If xm occurs in the expansion of (x+1x2)2n, then the coefficient of xm is

A
(2n)!(m)! (2nm)
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B
(2n)! 3! 3!(2nm)!
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C
(2n)!(2nm3)!(4n+m3)!
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D
(2n)!(2nm3)!(3n+m2)!
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Solution

The correct option is C (2n)!(2nm3)!(4n+m3)!
Tr+1= 2nCrx2nr(1x2)r= 2nCrx2n3r
This contains xm, If
2n3r=m
r=2nm3
Hence, coefficient of xm= 2nCr
=2n!(2nr)! r!

=2n!(2n2nm3)!(2nm3)!

=2n!(4n+m3)!(2nm3)!

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