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Find the middle term in the expansion of `(x^2+1/(x^2)-2)^ndot`

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Verified by Experts

The correct Answer is:
`((2n)!)/((n!)^(2))`
`(x^(2)+1/(x^(2))+2)^(n) = (x+1/x)^(2n)`
Hence, the middle term is
`T_((2n)/(2)+1) = T_(n+1) = .^(2n)C_(n)(x)^(n)(1/x)^(n) = .^(2n)C_(n)= ((2n)!)/((n!)^(2))`
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