Question

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

• A. $k=\pm 3$
• B. $k=\pm 5$
• C. $k=\pm 2$
• D. None of these

Answer 1

The correct option is A $k=\pm 3$

Given expression is $(\sqrt{x}-K/x^2{)}^{10}$

$T_{r+1}{=}^{10}{C}_r(\sqrt{2}{)}^{10-r}{\left(\frac{-k}{x^2}\right)}^r$

${=}^{10}{C}_r\left(x{)}^{1/2(10-r)}\right(-k{)}^{r\left(x^{-2r}\right)}$

${=}^{10}{C}_r\left(x{)}^{5-\frac{r}{2}-2r}\right(-k{)}^r{=}^{10}{C}_r{x}^{\frac{10-5r}{2}}(-k{)}^r$

For the term free from $x,\frac{10-5r}{2}=0\Rightarrow r=2$

So, the term free form $x$ is $T_{2+1}{=}^{10}{C}_2(-k{)}^2$

$\Rightarrow {}^{10}{C}_2(-k{)}^2=405$

$\Rightarrow \frac{10\times 9\times 8!}{2\times 8!}{k}^2=405$

$\Rightarrow k^2=9$

$\therefore k\pm 3$