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)]!)` .

Explanation

`(x^{2}+frac{1}{x})^{2 n}`

`T_{r+1}=\ ^{2 n} C r(x^{2})^{2 n-r}(frac{1}{x})^{r}`

`T_{r+1}=\ ^{2 n} C r x^{4 n-2 r-r}`

`T_{r+1}=\ ^{2 n} C r x^{4 n-3 r}`

`4 n-3 r=p`

`Rightarrow 4 n-p=3 r`

`Rightarrow r=frac{4 n-p}{3}`

`therefore` coefficient of `x^{p}` is `\ ^{2 m} C_{4 n-p} frac{4}{3}`

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