Thank you for reporting, we will resolve it shortly
Q.
Find the middle term in the expansion of $\left(2ax-\frac{b}{x^{2}}\right)^{12}$.
Binomial Theorem
Solution:
Since the power of the given expansion is even, it has one middle term which is $\left(\frac{12+2}{2}\right)^{th} = 7^{th}$ term
$T_{7 }= \,{}^{12}C_{6}\left(2ax\right)^{6}\left(\frac{-b}{x^{2}}\right)^{6}$
$ =\,{}^{12}C_{6} \frac{2^{6}\,a^{6}\,x^{6}\cdot\left(-b\right)^{6}}{x^{12}}$
$= \,{}^{12}C_{6} \frac{2^{6}\,a^{6}\,b^{6}}{x^{6}}$
$= \frac{59136 \,a^{6}\,b^{6}}{x^{6}}$