The middle term in the expansion of

$\left(x-\frac{1}{x}\right)^{18}$

Since 18 is even

there is only one middle term i.e. $\left(\frac{2 n}{2}+1\right)^{\text {th }}$ term

i.e. (n + 1)th term

i.e. $T_{n+1}={ }^{2 n} C_{n}(x)^{2 n-n}\left(\frac{-1}{x}\right)^{n}$

$={ }^{2 n} C_{n}(-1)^{n} x^{2 n-n} x^{-n}$

$={ }^{2 n} C_{n}(-1)^{n} x^{0} \quad$ i. e. ${ }^{18} C_{9}(-1)^{18}={ }^{18} C_{9}$