If the constant term in the binomial expansio | vedclass

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Question

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

Vedclass · Accepted Answer

$3$

Vedclass · Answer

$\left(\sqrt{x}-\frac{k}{x^{2}}\right)^{10}$

$T_{r+1}={ }^{10} C_{r}(\sqrt{x})^{10-r}\left(\frac{-k}{x^{2}}\right)^{r}$

$T_{r+1}={ }^{10} C_{r} \cdot x^{\frac{10-r}{2}} \cdot(-k)^{r} \cdot x^{-2 r}$

$T_{r+1}={ }^{10} C_{r} x^{\frac{10-5 r}{2}}(-k)^{r}$

Constant term $: \frac{10-5 r}{2}=0 \Rightarrow r=2$

$T_{3}={ }^{10} C_{2} \cdot(-k)^{2}=405$

$k^{2}=\frac{405}{45}=9$

$k=\pm 3 \Rightarrow|k|=3$

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

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