Question

In ${(\sqrt{x}-\frac{2}{x})}^{18}$, the term independent of $x$ is

• A. ${}^{18}{C}_6$
• B. $2x^{18}{C}_6$
• C. ${}^{18}{C}_6{.2}^6$
• D. $16x^{18}{C}_6$

Answer 1

Answer: (C)

${}^{18}{C}_6{.2}^6$

We know, $T_{r+1}={}^n{C}_r{a}^{n-r}{b}^r$
Applying to the above question, we get
$T_{r+1}=(-1){}^r{}^{18}{C}_r{x}^{\frac{18-r}{2}}{2}^r{x}^{-r}$
$=(-1){}^r{}^{18}{C}_r{2}^r{x}^{\frac{18-3r}{2}}$ ...(i)
For term independent of $x$
$\Rightarrow \frac{18-3r}{2}=0$
$\Rightarrow 18=3r$
$\Rightarrow r=6$
Substituting in (i), we get
$T_{6+1}=(-1){}^6{}^{18}{C}_6{2}^6$
$T_7={}^{18}{C}_6{2}^6$