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

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(C) The general term $T_{r+1}$ in the expansion of $\left(\frac{x}{4}-\frac{12}{x^{2}}\right)^{12}$ is given by:$T_{r+1} = {}^{12}C_{r} \left(\frac{x}

Vedclass · Accepted Answer

$55$

Vedclass · Answer

(C) The general term $T_{r+1}$ in the expansion of $\left(\frac{x}{4}-\frac{12}{x^{2}}ight)^{12}$ is given by:$T_{r+1} = {}^{12}C_{r} \left(\frac{x}{4}ight)^{12-r} \left(-\frac{12}{x^{2}}ight)^{r}$$T_{r+1} = {}^{12}C_{r} \left(\frac{1}{4}ight)^{12-r} (-12)^{r} x^{12-r} x^{-2r}$$T_{r+1} = {}^{12}C_{r} \left(\frac{1}{4}ight)^{12-r} (-1)^{r} (12)^{r} x^{12-3r}$For the term to be independent of $x$,the exponent of $x$ must be $0$:$12-3r = 0 \Rightarrow r = 4$Substituting $r=4$ into the expression:$T_{5} = {}^{12}C_{4} \left(\frac{1}{4}ight)^{8} (-1)^{4} (12)^{4}$$T_{5} = \frac{12 	imes 11 	imes 10 	imes 9}{4 	imes 3 	imes 2 	imes 1} 	imes \frac{1}{4^{8}} 	imes 12^{4}$$T_{5} = 495 	imes \frac{1}{4^{8}} 	imes (3 	imes 4)^{4} = 495 	imes \frac{3^{4} 	imes 4^{4}}{4^{8}} = 495 	imes \frac{3^{4}}{4^{4}}$Given the term is $\left(\frac{3^{6}}{4^{4}}ight) k$:$495 	imes \frac{3^{4}}{4^{4}} = \frac{3^{6}}{4^{4}} 	imes k$$k = \frac{495 	imes 3^{4}}{3^{6}} = \frac{495}{3^{2}} = \frac{495}{9} = 55$

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

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