If the constant term in the binomial expansion of

Question:

If the constant term in the binomial expansion of $\left(\sqrt{x} \frac{k}{x^{2}}\right)^{10}$ is 405, then $|k|$ equals:

  1. (1) 9

  2. (2) 1

  3. (3) 3

  4. (4) 2


Correct Option: , 3

Solution:

General term $=T_{r+1}={ }^{10} C_{r}(\sqrt{x})^{10-r} \cdot\left(-\frac{k}{x^{2}}\right)^{\eta}$

$={ }^{10} C_{r}(-k)^{r} \cdot x^{\frac{10-r}{2}-2 r}={ }^{10} C_{r}(-k)^{r} \cdot x^{\frac{10-5 r}{2}}$

Since, it is constant term, then

$\frac{10-5 r}{2}=0 \Rightarrow r=2$

$\therefore{ }^{10} C_{2}(-k)^{2}=405$

$\Rightarrow k^{2}=\frac{405 \times 2}{10 \times 9}=\frac{81}{9}=9$

$\therefore|k|=3$

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