Question

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

Answer 1

Given expression is `(x^2 + 1/x)^(2n)`

General terms, `"T"_(r + 1) = ""^n"C"_rx^(n - r) y^r`

= `""^(2n)"C"_r (x^2)^(2n - r) * (1/x)^r`

= `""^(2n)"C"_r (x)^(4n - 2r) * 1/x^r`

= `""^(2n)"C"_r (x)^(4n - 2r - r)`

= `""^(2n)"C"_r(x)^(4n - 3r)`

If x^p occurs in `(x^2 + 1/x)^(2n)`

Then 4n – 3r = p

⇒ 3r = 4n – p

⇒ r = `(4n - p)/3`

∴ Coefficient of x^p = `""^(2n)"C"_r = ""^(""2n)"C"_((4n - p)/3)`

= `((2n)!)/(((4n - p)/3)!(2n - (4n - p)/3)!)`

= `((2n)!)/(((4n - p)/3)!((6n - 4n + p)/3)!)`

= `((2n)!)/(((4n - p)/3)!((2n + p)/3)!)`

Hence, the coefficient of x^p = `((2n)!)/(((4n - p)/3)!((2n + p)/3)!)`