Question

The term independent of $x$ in the expansion of ${(\sqrt{x}-\frac{2}{\sqrt{x}})}^{18}$ is

• A. $-\left(\begin{array}{c}18\\ 9\end{array}\right)2^9$
• B. $\left(\begin{array}{c}18\\ 9\end{array}\right)2^{12}$
• C. $\left(\begin{array}{c}18\\ 9\end{array}\right)2^6$
• D. $\left(\begin{array}{c}18\\ 9\end{array}\right)2^8$

Answer 1

The correct option is A $-\left(\begin{array}{c}18\\ 9\end{array}\right)2^9$

${(\sqrt{x}-\frac{2}{\sqrt{x}})}^{18}$

General Term for the above binomial expansion will be

$T_{r+1}=(-1{)}^r{}^{18}{C}_r{x}^{\frac{18-r}{2}}.x^{-\frac{r}{2}}{2}^r$

$=(-1{)}^r{}^{18}{C}_r{x}^{\frac{18-2r}{2}}{.2}^r$

$=(-1{)}^r{}^{18}{C}_r{x}^{9-r}{.2}^r$

For the term independent of $x$, the power of $x$ has to be $0$.

Hence $9-r=0$

$\therefore r=9$

Hence, $T_{10}=-{}^{18}{C}_9{2}^9$