How do you find coefficient of specific x term in binomial expansion

Question

How do you find the term in $x^5$ in the expansion of $((1/x)+x)^9$?

Answer 1

By the binomial theorem, $$ (y+x)^9=\sum_{k=0}^{9}\binom{9}{k}y^kx^{9-k} $$ For $y=x^{-1}$ this becomes $$ \sum_{k=0}^{9}\binom{9}{k}x^{-k}x^{9-k}= \sum_{k=0}^{9}\binom{9}{k}x^{9-2k} $$ Can you finish?

The only way to get $x^5$ is for $9-2k=5$, that is, $k=2$, so the coefficient is $\dbinom{9}{2}=36$.