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Q.
Find the term independent of $x$ in the expansion of $\left(\frac{\sqrt{x}}{\sqrt{3}}+\frac{\sqrt{3}}{2x^{2}}\right)^{10}$.
Binomial Theorem
Solution:
Let $(r + 1)^{th}$ term be independent of $x$.
$T_{r+1} = \,{}^{10}C_{r}\left(\sqrt{\frac{x}{3}}\right)^{10-r}\left(\frac{\sqrt{3}}{2x^{2}}\right)^{r}$
$= \,{}^{10}C_{r}\left(\frac{x}{3}\right)^{\frac{10-r}{2}} \,3^{\frac{r}{2}}\left(\frac{1}{2^{r}\,x^{2r}}\right)$
$= \,{}^{10}C_{r}3^{\frac{r}{2}-\frac{10-r}{2}}\,2^{-r}\,x^{\frac{10-r}{2}-2r}$
Since the term is independent of $x$, we have
$\frac{10-r}{2}-2r=0$
$\Rightarrow r =2$
$\therefore T_{3} = \,{}^{10}C_{2} \frac{3^{-3}}{4}$
$= \frac{10 \times 9}{2 \times 1}\times \frac{1}{9 \times 12} = \frac{5}{12}$