Question

The term independent of $x$ in the expansion of ${\left(\sqrt{\frac{x}{3}}+\frac{3}{2x^2}\right)}^{10}$ is

• A. $\frac{5}{4}$
• B. $\frac{7}{4}$
• C. $\frac{9}{4}$
• D. $45$

Answer 1

Answer: (A)

General term, $T_{r+1}{=}^{10}{C}_r{\left(\sqrt{\frac{x}{3}}\right)}^{10-r}.{\left(\frac{3}{2x^2}\right)}^r$
${=}^{10}{C}_r{3}^{r-\frac{10-r}{2}}{\left(\frac{1}{2}\right)}^r.x^{\frac{10-r}{2}-2r}$
${=}^{10}{C}_r{3}^{r-\frac{3r-10}{2}}{\left(\frac{1}{2}\right)}^r.x^{\frac{10-5r}{2}}$
For the term independent of x, put
$\frac{10-5r}{2}=0$
$\Rightarrow$ $5r=10$
$\Rightarrow$ $r=2$
$\therefore$ The term independent of x,
$T_3{=}^{10}{C}_2{3}^{\frac{6-10}{2}}{\left(\frac{1}{2}\right)}^2$ $=\frac{10\times 9}{2\times 3^2\times 4}=\frac{5}{4}$