The coefficient of the term independent of x | vedclass

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(A) The $(r+1)^{th}$ term in the expansion of $\left(\sqrt{\frac{x}{3}}+\frac{3}{2 x^{2}}\right)^{10}$ is given by$T_{r+1} = {^{10}C_r} \left(\sqrt{\f

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$5/4$

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(A) The $(r+1)^{th}$ term in the expansion of $\left(\sqrt{\frac{x}{3}}+\frac{3}{2 x^{2}}ight)^{10}$ is given by$T_{r+1} = {^{10}C_r} \left(\sqrt{\frac{x}{3}}ight)^{10-r} \left(\frac{3}{2x^2}ight)^r$$= {^{10}C_r} \frac{x^{(10-r)/2}}{3^{(10-r)/2}} \cdot \frac{3^r}{2^r x^{2r}}$$= {^{10}C_r} \frac{3^{r - (10-r)/2}}{2^r} x^{(10-r)/2 - 2r}$For the term to be independent of $x$,the exponent of $x$ must be zero:$\frac{10-r}{2} - 2r = 0 \implies 10 - r - 4r = 0 \implies 5r = 10 \implies r = 2$Substituting $r=2$ into the coefficient expression:Coefficient $= {^{10}C_2} \cdot \frac{3^{2 - (10-2)/2}}{2^2} = {^{10}C_2} \cdot \frac{3^{2-4}}{4} = \frac{10 	imes 9}{2} \cdot \frac{3^{-2}}{4} = 45 \cdot \frac{1}{9 	imes 4} = \frac{45}{36} = \frac{5}{4}$

The coefficient of the term independent of $x$ in the expansion of ${\left( {\sqrt {\frac{x}{3}} + \frac{3}{{2{x^2}}}} \right)^{10}}$ is

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