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Question

If xp occurs in the expansion of (x2+1x)2n, prove that its coefficient is ⎢ ⎢ ⎢ ⎢(2n)!(4np3)!(2n+p3)!⎥ ⎥ ⎥ ⎥

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Solution

(x2+1x)2n
Tr+1=2nCr(x2)2nr(1x)r
Tr+1=2nCrx4n2rr
Tr+1=2nCrx4n3r
4n3r=p
4np=3r
r=4np3
coefficient of xp is 2nC4np3
=(2n)!(4np3)!(2n+p3)!

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