Show that the middle term in the expansion of

To show: that the middle term in the expansion of $\left(\frac{2 x^{2}}{3}+\frac{3}{2 x^{2}}\right)^{10}$ is 252.

Formula Used:

General term, $T_{r+1}$ of binomial expansion $(x+y)^{n}$ is given by,

$T_{r+1}={ }^{n} C_{r} x^{n-r} y^{r}$ where

${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=\frac{n !}{r !(n-r) !}$

Total number of terms in the expansion is 11

Thus, the middle term of the expansion is $T_{6}$ and is given by,

$\mathrm{T}_{6}={ }^{10} \mathrm{C}_{5} \times\left(\frac{2 x^{2}}{3}\right)^{5} \times\left(\frac{3}{2 \mathrm{x}^{2}}\right)^{5}$

$\mathrm{T}_{6}={ }^{10} \mathrm{C}_{5}$

$T_{6}=\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 5 \times 4 \times 3 \times 2}$

$T_{6}=252$

Thus, the middle term in the expansion of $\left(\frac{2 x^{2}}{3}+\frac{3}{2 x^{2}}\right)^{10}$ is $252 .$