Question

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)]!)` .

Answer 1

Given expansion is `(x^(2) + (1)/(x))^(2n)`
Let `x^(p)` occur in the expansion of `(x^(2) + (1)/(x))^(1n)`
`T_(r + 1) = .^(2n)C_(r) (x^(2))^(2n - r) ((1)/(x))^(r)`
`= .^(2n)C_(r) x^(4n - 2n) x^(-r) = .^(2n)C_(r) x^(4n - 3r)`
Let `4n - 3r = p`
`:.` Coefficient of `x^(p) = .^(2n)C_(r) = ((2n)!)/(r! (2n - r)1) = ((2n)!)/(((4 n - p)/(3)) ! (2n - (4n - p)/(3))!)`
`= ((2n)!)/(((4 n - p)/(3))! ((6 n - 4n - p)/(3))!) = ((2n)!)/(((4n - p)/(3))! ((2n + p)/(3))!)`