Question

In the expansion of ${\left(\begin{array}{c}\sqrt{x}+\frac{1}{3x^2}\end{array}\right)}^{10}$ the value of constant term (independent of $x$) is

• A. 5
• B. 8
• C. 45
• D. 90

Answer 1

The correct option is A 5

Solution:

We have, ${(\sqrt{x}+\frac{1}{3x^2})}^{10}$

$\therefore T_{r+1}={}^{10}{C}_r{x}^{-2r}{x}^{\frac{10-r}{2}}{3}^{-r}$$={}^{10}{C}_r{x}^{\frac{10-5r}{2}}{3}^{-r}$

For the term independent of $x,$

$\frac{10-5r}{2}=0$

or, $r=2$

Put, $r=2$, we get

$T_3={}^{10}{C}_2(3{)}^{-2}=\frac{10\times 9}{1\times 2}\times \frac{1}{9}=5$

Hence, A is the correct option.