The coefficient of the term independent of x in the expansion of (...

The $(r+1)$ th term in the expansion of \left(\sqrt{\frac{x}{3}}+\frac{3}{2 x^{2}}\right)^{10} is given by $$\begin{array}{l} \mathrm{T}_{r+1}={ }^{10} \mathrm{C}_{r}\left(\sqrt{\frac{x}{3}}\right)^{10-r}\left(\frac{3}{2 x^{2}}\right)^{r} \\ ={ }^{10} \mathrm{C}_{r} \frac{x^{5-\left(\frac{r}{2}\right)}}{3^{5-\left(\frac{r}{2}\right)}} \cdot \frac{3^{r}}{2^{r} x^{2 r}} \\ ={ }^{10} \mathrm{C}_{r} \frac{3^{\left(\frac{3 r}{2}\right)-5}-5-\left(\frac{5 r}{2}\right)}{2^{r}} x \end{array}$$\mathrm{T}_{r+1}={ }^{10} \mathrm{C}_{r}\left(\sqrt{\frac{x}{3}}\right)^{10-r}\left(\frac{3}{2 x^{2}}\right)^{r}={ }^{10} \mathrm{C}_{r} \frac{x^{5-\left(\frac{r}{2}\right)}}${3} \cdot \frac{3^{r}}{2^{r} x^{2 r}}={ }^{10} \mathrm{C}_{r} \frac{3^{\left(\frac{3 r}{2}\right)-5}-\left(\frac{5}{2}\right)-\left(\frac{5 r}{2}\right)}{2^{r}}$x$For \mathrm{T}_{r+1} to be independent of$x$, we must have$5-\left(\frac{5 r}{2}\right)=$0$ or $r=2$. Thus, the 3 rd term is independent of $x$ and is equal to ${ }^{10}$\mathrm{C}_{2} \frac{3^{3-5}}{2^{2}}=\frac{10 \times 9}{2} \times \frac{3^{-2}}{4}=\frac{5}{4}$