The coefficient of x^2 term in the binomial expansion of ((1)/(3)...

General term of $given binomial expansion is$$\begin{aligned} T_{r+1} & ={ }^{10} C_r\left(\frac{1}{3} x^{\frac{1}{2}}\right)^{10-r}\left(x^{-\frac{1}{4}}\right)^r \\ & ={ }^{10} C_r \times\left(\frac{1}{3}\right)^{10-r} x^{\frac{10-r}{2}-\frac{r}{4}} \end{aligned}$We have to find coefficient of $x^2\begin{aligned} \frac{10-r}{2}-\frac{r}{4} & =2 \Rightarrow r=4 \\ \Rightarrow \quad T_{4+1} & ={ }^{10} C_4\left(\frac{1}{3}\right)^6 x^2 \\ \therefore \text{ Coefficient of } x^2 & ={ }^{10} C_4\left(\frac{1}{3}\right)^6 \\ & =\frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} \times \frac{1}{3^6}=\frac{70}{243} \end{aligned}$$$