## College Algebra 7th Edition

$48620x^{18}$
We find the middle term (9th power) in the expansion of $(x^{2}+1)^{18}$ by using Binomial Theorem coefficients: $\left(\begin{array}{l}18\\9\end{array}\right)(x^{2})^{9}(1)^{9}=48620(x^{18})(1)^9=48620x^{18}$