Question

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

• A. $\frac{45}{64}$
• B. $\frac{64}{45}$
• C. $\frac{6}{5}$
• D. $\frac{4}{5}$

Answer 1

Answer: (A)

$\frac{45}{64}$

We know, $T_{r+1}={}^n{C}_r{a}^{n-r}{b}^r$
Applying to the above question, we get
$T_{r+1}=(-1){}^r{}^{10}{C}_r{2}^{r-10}{x}^{\frac{10-r}{2}}{2}^r{x}^{-2r}$
$=(-1){}^r{}^{10}{C}_r{2}^{2r-10}{x}^{\frac{10-5r}{2}}$ ...(i)
For term independent of $x$
$\Rightarrow \frac{10-5r}{2}=0$
$\Rightarrow 10=5r$
$\Rightarrow r=2$
Substituting in equation (i), we get
$T_{2+1}=(-1){}^2{}^{10}{C}_2{2}^{-6}$
$T_3={}^{10}{C}_2{2}^{-6}$ $=\frac{10.9}{2!{.2}^6}$ $=\frac{45}{64}$