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प्रश्न
If xp occurs in the expansion of `(x^2 + 1/x)^(2n)`, prove that its coefficient is `(2n!)/(((4n - p)/3)!((2n + p)/3)!)`
उत्तर
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 xp occurs in `(x^2 + 1/x)^(2n)`
Then 4n – 3r = p
⇒ 3r = 4n – p
⇒ r = `(4n - p)/3`
∴ Coefficient of xp = `""^(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 xp = `((2n)!)/(((4n - p)/3)!((2n + p)/3)!)`
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