Find the middle term(s) in the expansion of :
$\left(2x-\dfrac{x^{2}}{4}\right)^{9}$

Given expansion term is $\left(2x-\dfrac{x^{2}}{4}\right)^{9}$

$n=9=$ odd

middle term $=\left(\dfrac{9+1}{2}\right)^{th}+\left(\dfrac{9+3}{2}\right)^{th}$

$=5^{th}$ or $6^{th}$ term

$T_{5}=T_{4+1}=\ ^{9}C_{4}(2x)^{5}\left(\dfrac{-x^{2}}{4}\right)^{4}$

$=\dfrac{9!}{4!5!}2^{5}\times x^{5}(-1)^{4}\times \dfrac{x^{8}}{4^{4}}$

$=\dfrac{9\times 8\times 7\times 6\times 5!}{4\times 3\times 2\times 5!}\times \dfrac{2^{5}.x^{13}.1}{2^{5}}$

$T_{5}=\dfrac{63}{4}x^{13}$

$T_{6}=T_{5+1}=\ ^{9}C_{5}(2x)^{4}\left(\dfrac{-x^{2}}{4}\right)^{5}$

$=\dfrac{9!}{4!5!}\times 2^{4}x^{4}(-1)^{5}\dfrac{x^{10}}{4^{5}}$

$=\dfrac{9\times 8\times 7\times 6\times 5!}{4\times 3\times 2\times 5!}2^{4}\dfrac{x^{14}}{2^{10}}$

$T_{6}=\dfrac{-63}{32}x^{14}$