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(A) The general term in the expansion of $\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{6}$ is given by $T_{r+1} = {}^{6}C_{r} \left(\frac{3}{2} x^{2}

Vedclass · Accepted Answer

$\frac{5}{12}$

Vedclass · Answer

(A) The general term in the expansion of $\left(\frac{3}{2} x^{2}-\frac{1}{3 x}ight)^{6}$ is given by $T_{r+1} = {}^{6}C_{r} \left(\frac{3}{2} x^{2}ight)^{6-r} \left(-\frac{1}{3x}ight)^{r}$.$T_{r+1} = {}^{6}C_{r} \left(\frac{3}{2}ight)^{6-r} (x^{2})^{6-r} (-1)^{r} \left(\frac{1}{3}ight)^{r} (x^{-1})^{r}$.$T_{r+1} = {}^{6}C_{r} \left(\frac{3}{2}ight)^{6-r} (-1)^{r} \left(\frac{1}{3}ight)^{r} x^{12-2r-r}$.$T_{r+1} = {}^{6}C_{r} \frac{3^{6-r}}{2^{6-r}} \frac{(-1)^{r}}{3^{r}} x^{12-3r}$.For the term to be independent of $x$,the exponent of $x$ must be $0$,so $12-3r = 0$,which gives $r = 4$.Substituting $r=4$ into the expression:$T_{4+1} = {}^{6}C_{4} \frac{3^{6-4}}{2^{6-4}} \frac{(-1)^{4}}{3^{4}} = {}^{6}C_{2} \frac{3^{2}}{2^{2}} \frac{1}{3^{4}} = 15 	imes \frac{9}{4} 	imes \frac{1}{81} = 15 	imes \frac{1}{4 	imes 9} = \frac{15}{36} = \frac{5}{12}$.

Find the term independent of $x$ in the expansion of $\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{6}$.

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