Q95 / 160 · TS EAMCET 2014 · Mathematics
The term independent of $x$ in the expansions of $\left(\sqrt{x}-\frac{2}{\sqrt{x}}\right)^{18}$ is
- Correct answerA$-{ }^{18} C_9 2^9$
- B${ }^{18} C_9 2^{12}$
- C${ }^{18} C_6 2^6$
- D${ }^{18} C_6 2^8$
Solution
General term in the expansion of $\left(\sqrt{x}-\frac{2}{\sqrt{x}}\right)^{18}$ is
$$
\begin{aligned}
T_{r+1} & ={ }^{18} C_r(\sqrt{x})^{18-r}\left(-\frac{2}{\sqrt{x}}\right)^r \\
& ={ }^{18} C_r(x)^{\frac{18-r}{2}}(-2)^r(x)^{-\frac{r}{2}} \\
& ={ }^{18} C_r(-2)^r x^{9-r}
\end{aligned}
$$
For independent of $x$,
$$
\begin{array}{rlrl}
& \text { put } 9-r & =0 \\
\Rightarrow & & r & =9 \\
\therefore & & T_{10} & ={ }^{18} C_9(-2)^9 \\
& =-{ }^{18} C_9 2^9
\end{array}
$$
Hence, the correct answer is option: 'A'.