Q.Find the middle term in the expansion of $\left(x + \dfrac{1}{x}\right)^{8}$.

Since $n=8$ is even, there is a single middle term, the $\left(\dfrac{8}{2}+1\right)$th $=$ 5th term, found from the general term formula.

$(x+1/x)^8$ has $8+1=9$ terms. Since $n=8$ is even, there is a single middle term, the $\left(\dfrac{n}{2}+1\right)=\left(\dfrac{8}{2}+1\right)=5$th term.

Using the general term $T_{r+1} = \,^{8}C_{r}\,x^{8-r}\left(\dfrac{1}{x}\right)^{r} = \,^{8}C_{r}\,x^{8-2r}$, the middle term is at $r=4$:

$^{8}C_{4} = \dfrac{8!}{4!\,4!} = 70$.

So the middle term is $70$. …

Using the odd-$n$ rule (two middle terms) when $n=8$ is actually even; also forgetting that $\left(\dfrac{1}{x}\right)^r = x^{-r}$, which is …