Question

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

Answer 1

The $(r+1{)}^{th}$ term is $T_{r+1}{=}^{10}{C}_r\left(\sqrt{\frac{x}{3}}{)}^{10-r}\right(\frac{\sqrt{3}}{2x^2}{)}^r$

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

$\therefore T_{r+1}{=}^{10}{C}_r\frac{\sqrt{3}}{3^{\frac{10-r}{2}}\times 2^r}{x}^{\frac{10-r}{2}-2r}$ ........(i)
The term becomes independent when exponent of x is zero.

$\Rightarrow \frac{10-r}{2}-2r=0\Rightarrow \frac{10-r}{2}=2r\Rightarrow 10-r=4r$

$\Rightarrow 10=5r\Rightarrow r=2$
Now, substitute $r=2$ in equation (i), we get

$T_3{=}^{10}{C}_2\frac{(\sqrt{3}{)}^2}{3^4{.2}^2}=\frac{10!}{8!2!}\times \frac{3}{3^4{2}^2}$

$T_3=\frac{10\times 9}{2\times 3\times 3\times 3\times 4}$

Hence, Independent term, $T_3=\frac{5}{12}$