Mathematics - Binomial Theorem Question with Solution | TestHub

General term of given binomial expansion is $\begin{array}{rl}{T}_{r+1}& ={}^{10}{C}_r{\left(\frac{1}{3}{x}^{\frac{1}{2}}\right)}^{10-r}{\left(x^{-\frac{1}{4}}\right)}^r\\ & ={}^{10}{C}_r\times {\left(\frac{1}{3}\right)}^{10-r}{x}^{\frac{10-r}{2}-\frac{r}{4}}\end{array}$ We have to find coefficient of $x^2$ $\begin{array}{rl}\frac{10-r}{2}-\frac{r}{4}& =2\Rightarrow r=4\\ \Rightarrow T_{4+1}& ={}^{10}{C}_4{\left(\frac{1}{3}\right)}^6{x}^2\\ \therefore \text{ Coefficient of }{x}^2& ={}^{10}{C}_4{\left(\frac{1}{3}\right)}^6\\ & =\frac{10\times 9\times 8\times 7}{4\times 3\times 2\times 1}\times \frac{1}{3^6}=\frac{70}{243}\end{array}$