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

The general term in the expansion of `(x^(2) + 1/ x) ^(2n) ` is given by
`T _(r+1) = .^(2n) C _(r) xx (x^(2))^((2n-r)) xx (1/x)^(r)`
`rArr T _ (r+1) = .^(2n)C_(r) xx x ^((4n-3r))." "` ...(i)
This term contains `x^(p) " only when " 4n- 3r = p .`
And, `4n- 3r = p rArr ((4n - p))/3.`
Putting `4n-3r=p` in (i), we get
coefficent of `x^(p) = .^(2n) C _ ( r) , " where " r = ((4n-p))/3`
`((2n) !)/((r!) xx (2n- r) !)=((2n))/({((4n-p)/3)! } xx {[2n-((4n-p))/3]!})`
`((2n)!)/({((4n-p)/3)!} xx {((2n+p)/3)!}).`
Hence , the coefficent of `x^(p) " in the expansion of " (x^(2) + 1/x)^(2n)` is
`((2n)!)/({((4n-p))/3!} xx { ( (2n+p)/3)!}).`