Question

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

• A. $3$
• B. $5$
• C. $9$
• D. $Noneofthese$

Answer 1

Answer: (C)

$9$

As we know that the general term in an expansion ${(a+b)}^n$ is given as-

$T_{r+1}={}^n{C}_r{a}^r{b}^{n-r}$

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

Here, $a=\sqrt{\frac{x}{3}},b=\frac{3}{2x^2},n=10$

Therefore,$T_{r+1}={}^{10}{C}_r{\left(\sqrt{\frac{x}{3}}\right)}^r{\left(\frac{3}{2x^2}\right)}^{10-r}$

$\Rightarrow T_{r+1}={}^{10}{C}_r{x}^{(\frac{r}{2}-20+2r)}{3}^{(10-r-\frac{r}{2})}{2}^{(r-10)}={}^{10}{C}_r{x}^{(\frac{5r}{2}-20)}{3}^{(10-\frac{3r}{2})}{2}^{(r-10)}$

For the term independent of $x$ is $\frac{5r}{2}-20=0$
$\Rightarrow \frac{5r}{2}=20$

$\Rightarrow r=\frac{20\times 2}{5}=8$

Hence, ${(8+1)}^{th}=9^{th}$ term will be independent of $x$.