Mathematics - Binomial Theorem Question with Solution | TestHub

The $(r+1)$ th term in the expansion of ${\left(\sqrt{\frac{x}{3}}+\frac{3}{2x^2}\right)}^{10}$ is given by $\begin{array}{l}{\mathrm{T}}_{r+1}={}^{10}{\mathrm{C}}_r{\left(\sqrt{\frac{x}{3}}\right)}^{10-r}{\left(\frac{3}{2x^2}\right)}^r\\ ={}^{10}{\mathrm{C}}_r\frac{x^{5-\left(\frac{r}{2}\right)}}{3^{5-\left(\frac{r}{2}\right)}}\cdot \frac{3^r}{2^r{x}^{2r}}\\ ={}^{10}{\mathrm{C}}_r\frac{3^{\left(\frac{3r}{2}\right)-5}-5-\left(\frac{5r}{2}\right)}{2^r}x\end{array}$ ${\mathrm{T}}_{r+1}={}^{10}{\mathrm{C}}_r{\left(\sqrt{\frac{x}{3}}\right)}^{10-r}{\left(\frac{3}{2x^2}\right)}^r$ $={}^{10}{\mathrm{C}}_r\frac{x^{5-\left(\frac{r}{2}\right)}}{3}\cdot \frac{3^r}{2^r{x}^{2r}}$ $={}^{10}{\mathrm{C}}_r\frac{3^{\left(\frac{3r}{2}\right)-5}-\left(\frac{5}{2}\right)-\left(\frac{5r}{2}\right)}{2^r}x$ For ${\mathrm{T}}_{r+1}$ to be independent of $x$, we must have $5-\left(\frac{5r}{2}\right)=0$ or $r=2$. Thus, the 3 rd term is independent of $x$ and is equal to ${}^{10}{\mathrm{C}}_2\frac{3^{3-5}}{2^2}=\frac{10\times 9}{2}\times \frac{3^{-2}}{4}=\frac{5}{4}$