Question

If the term free from $x$ in the expansion of ${\left(\sqrt{x}-\frac{k}{x^2}\right)}^{10}$ is $405$, then the value of $k$ is:

• A. $\pm 1$
• B. $\pm 3$
• C. $\pm 4$
• D. $\pm 2$

Answer 1

The correct option is B

$\pm 3$

Explanation for the correct option.

Step 1: Find the value of $r$.

The general term for ${\left(a+b\right)}^n$ is $T_{r+1}={}^{n}C_r{a}^{n-r}{b}^r$.

So for ${\left(\sqrt{x}-\frac{k}{x^2}\right)}^{10}$,

$\begin{array}{rcl}{T}_{r+1}& =& {}^{10}C_r{\left(\sqrt{x}\right)}^{10-r}{\left(\frac{-k}{x^2}\right)}^r\\ & =& {}^{10}C_r{\left(x\right)}^{\frac{1}{2}\left(10-r\right)}\frac{{\left(-k\right)}^r}{x^{2r}}\\ & =& {}^{10}C_r{\left(x\right)}^{5-\frac{r}{2}-2r}{\left(-k\right)}^r\\ & =& {}^{10}C_r{\left(x\right)}^{\frac{10-5r}{2}}{\left(-k\right)}^r\end{array}$

It is given that the term is free from $x$, that means $\frac{10-5r}{2}=0$,

$\Rightarrow r=2$

Step 2: Find the value of $k$.

$\begin{array}{rcl}405& =& {}^{10}C_2{\left(-k\right)}^2\\ & \Rightarrow & 405=\frac{10\times 9\times 8!}{2\times 8!}\left(k^2\right)\\ & \Rightarrow & 405=45\left(k^2\right)\\ \Rightarrow k^2& =& 9\\ \Rightarrow k& =& \pm 3\end{array}$

Hence, option B is correct.