Find the middle term in the expansion of ${\left(2x^2-\frac{{\displaystyle 1}}{{\displaystyle x}}\right)}^7$

Here, $n=7$(odd)

$\therefore$There are two middle terms.

First middle term $={\left(\frac{{\displaystyle n+1}}{{\displaystyle 2}}\right)}^{th}={\left(\frac{{\displaystyle 7+1}}{{\displaystyle 2}}\right)}^{th}={\left(\frac{{\displaystyle 8}}{{\displaystyle 2}}\right)}^{th}=$4^th term

Second middle term $={\left(\frac{{\displaystyle n+3}}{{\displaystyle 2}}\right)}^{th}={\left(\frac{{\displaystyle 7+3}}{{\displaystyle 2}}\right)}^{th}={\left(\frac{{\displaystyle 10}}{{\displaystyle 2}}\right)}^{th}=$5^th term

$\begin{array}{l}\therefore T_4=T_{3+1}{=}^7{C}_3.{\left(2x^2\right)}^4.{\left(-\frac{1}{x}\right)}^3\\ ={-}^7{C}_3{.2}^4.x^5=\frac{-7!}{3!4!}\times 16x^5=-35\times 16x^5=-560x^5\end{array}$

And $T_5=T_{4+1}{=}^7{C}_4.{\left(2x^2\right)}^3.{\left(-\frac{{\displaystyle 1}}{{\displaystyle x}}\right)}^4$

${=}^7{C}_4{\left(2\right)}^3.x^2=\frac{7!}{4!3!}{\left(2\right)}^3.x^2=35\times 8\times x^2=280x^2$