Find the middle terms in the expansions of
Question:

Find the middle terms in the expansions of $\left(\frac{x}{3}+9 y\right)^{10}$

Solution:

It is known that in the expansion $(a+b)^{n}$, if $n$ is even, then the middle term is $\left(\frac{\mathrm{n}}{2}+1\right)^{\text {th }}$ term.

Therefore, the middle term in the expansion of $\left(\frac{x}{3}+9 y\right)^{10}$ is $\left(\frac{10}{2}+1\right)^{\text {th }}=6^{\text {th }}$ term

$\mathrm{T}_{6}=\mathrm{T}_{5+1}={ }^{10} \mathrm{C}_{5}\left(\frac{\mathrm{x}}{3}\right)^{10-5}(9 \mathrm{y})^{3}=\frac{10 !}{5 ! 5 !} \cdot \frac{\mathrm{x}^{5}}{3^{5}} \cdot 9^{5} \cdot \mathrm{y}^{5}$

$=\frac{10.9 \cdot 8 \cdot 7 \cdot 6.5 !}{5 \cdot 4 \cdot 3 \cdot 2.5 !} \cdot \frac{1}{3^{5}} \cdot 3^{10} \cdot x^{5} y^{5} \quad\left[9^{5}=\left(3^{2}\right)^{5}=3^{10}\right]$

$=252 \times 3^{5} \cdot x^{5} \cdot y^{5}=61236 x^{5} y^{5}$

Thus, the middle term in the expansion of $\left(\frac{x}{3}+9 y\right)^{10}$ is $61236 x^{5} y^{5}$.