Question

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

• A. $5/9$
• B. $5/3$
• C. $1/3$
• D. $4/3$

Answer 1

The correct option is B $5/3$

${(\sqrt{\frac{x}{3}}+\sqrt{\frac{\sqrt{3}}{x^2}})}^{10}$

$Trel{=}^{10}{C}_r{\left(\frac{\sqrt{x}}{\sqrt{3}}\right)}^{10-r}{\left(\frac{\sqrt{3}}{x^2}\right)}^r$

${=}^{10}{C}_r\times \frac{x^{\frac{10-r}{2}}}{(\sqrt{3}{)}^{10-r}}\times x^{-2r}\times (\sqrt{3}{)}^r$

${=}^{10}{C}_r\times x^{\frac{10-r}{2}-2r}\times \frac{(\sqrt{3}{)}^r}{{\sqrt{3}}^{10-r}}$

$\frac{10-r}{2}-2r=0$

$\frac{10-r}{2}-2r$

$10-r=4r$

$10=5r$

$r=2$

${}^{10}{C}_r\times \frac{(\sqrt{3}{)}^2}{\sqrt{3}}{=}^{10}{C}_r\times \frac{3}{3\times 3\times 3\times 3}$

${=}^{10}{C}_r\times \frac{1}{27}$

$=\frac{10!}{8!2!\times 27}=\frac{10\times 9\times 8!}{8!\times 2\times 27}$

$=\frac{90}{54}$

$=\frac{5}{3}$