निम्नलिखित प्रसारों में मध्य पद ज्ञात कीजिए

$\left(\frac{x}{3}+9 y\right)^{10}$

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

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

$ = \frac{{7 \cdot 6 \cdot 5 \cdot 4!}}{{4! \cdot 3 \cdot 2}} \cdot \frac{{{3^3}}}{{{2^4} \cdot {3^4}}} \cdot {x^{12}} = \frac{{35}}{{48}}{x^{12}}{T_4} = {T_{5 + 1}} = {\,^{10}}{C_5}{\left( {\frac{x}{3}} \right)^{10 - 5}}{(9y)^5} = \frac{{10!}}{{515!}} \cdot \frac{{{x^5}}}{{{3^5}}} \cdot {9^5} \cdot {y^5}$

$ = \frac{{10 \cdot 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}$            $\left[9^{5}=\left(3^{2}\right)^{5}=3^{10}\right]$

$ = 252 \times {3^5} \cdot {x^5} \cdot {y^5} = 6123{x^5}{y^5}$

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