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If x^p occurs in the expansion of (x^2+1...

If `x^p` occurs in the expansion of `(x^2+1/x)^(2n)` , prove that its coefficient is `((2n)!)/([1/3(4n-p)]![1/3(2n+p)]!)` .

Text Solution

Verified by Experts

The `(r+1)` th term is `.^(2n)C_(r)(x^(2))^(2n-r)(1//x)^(r ) = .^(2n)C_(r ) x^(4n-3r)`
For the coeficient of `x^(p)` ,
`4n -3r = p`
or `r = 1/3 (4n-p)`
Thererfore, the coefficient of `x^(p)` is
`.^(2n)C_((4n-p)//3)=((2n)!)/([1/3(4n-p)]![2n-1/3(4n-p)]!)`
`= ((2n)!)/([1/3(4n-p)]![(1)/(3)(2n+p)]!)`
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