If left(frac{3^{6}}{4^{4}}right) mathrm{k} is | vedclass

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Question

$\left(\frac{x}{4}-\frac{12}{x^{2}}\right)^{12}$

$T_{r+1}=(-1)^{r} \cdot{ }^{12} C_{r}\left(\frac{x}{4}\right)^{12-\mathrm{r}}\left(\frac{12}{x^{2}

Vedclass · Accepted Answer

$55$

Vedclass · Answer

$\left(\frac{x}{4}-\frac{12}{x^{2}}\right)^{12}$

$T_{r+1}=(-1)^{r} \cdot{ }^{12} C_{r}\left(\frac{x}{4}\right)^{12-\mathrm{r}}\left(\frac{12}{x^{2}}\right)^{r}$

$T_{r+1}=(-1)^{r} \cdot{ }^{12} C_{r}\left(\frac{1}{4}\right)^{12-r}(12)^{r} \cdot(x)^{12-3 r}$

Term independent of $x \Rightarrow 12-3 r=0 \Rightarrow r=4$

$\mathrm{T}_{5}=(-1)^{4} \cdot{ }^{12} \mathrm{C}_{4}\left(\frac{1}{4}\right)^{8}(12)^{4}=\frac{3^{6}}{4^{4}} \cdot \mathrm{k}$

$\Rightarrow \mathrm{k}=55$

If $\left(\frac{3^{6}}{4^{4}}\right) \mathrm{k}$ is the term, independent of $\mathrm{x}$, in the binomial expansion of $\left(\frac{\mathrm{x}}{4}-\frac{12}{\mathrm{x}^{2}}\right)^{12}$, then $\mathrm{k}$ is equal to ...... .

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