Question

find the term independent of x in the expansion of {√x/√3+√3/2x square} power 10..

Answer 1

Please see the attachment

Answer 2

$Given \: \Big( \sqrt{\frac{x}{3}} + \frac{\sqrt{3}}{2x^{2}} \Big)^{10}$

$We \:know \:that , \\General \: term \: in \:the \: expansion \: of \: (x+a)^{n} \: is :$

$\pink { t_{(r+1)} = ^{n}C_{r}\times x^{n-r} \times a^{r} }$

$Here , t_{(r+1)} = ^{10}C_{r}\times \Big(\sqrt{\frac{x}{3}}\Big)^{10-r} \times \Big( \frac{\sqrt{3}}{2x^{2}}\Big)^{r}$

$= ^{10}C_{r}\times \Big( \frac{1}{\sqrt{3}}\Big)^{10-r} \times \Big( \frac{\sqrt{3}}{2}\Big)^{r} \times (\sqrt{x})^{10-r} \times x^{-2r}$

$= ^{10}C_{r}\times \Big( \frac{1}{\sqrt{3}}\Big)^{10-r} \times \Big( \frac{\sqrt{3}}{2}\Big)^{r} \times (x)^{\frac{10-r}{2}-2r}$

$Now, (x)^{\frac{10-r}{2}-2r} = x^{0} \\\blue { (Term \: independent )}$

$\implies \frac{10-r}{2}-2r = 0$

$\implies \frac{10-r-4r}{2} = 0$

$\implies 10-5r= 0$

$\implies -5r= -10$

$\implies r=\frac{ -10}{-5}$

$\implies r = 2$

$The \: term \: independent \: of \: x \: in \:the \\expansion \: of \: \Big( \sqrt{\frac{x}{3}} + \frac{\sqrt{3}}{2x^{2}} \Big)^{10}$

$= t_{2+1}$

$= ^{10}C_{2}\times \Big( \frac{1}{\sqrt{3}}\Big)^{10-2} \times \Big( \frac{\sqrt{3}}{2}\Big)^{2}\\= ^{10}C_{2}\times \Big( \frac{1}{\sqrt{3}}\Big)^{8} \times \Big( \frac{\sqrt{3}}{2}\Big)^{2}\\= ^{10}C_{2}\times \frac{1}{3^{4}} \times \frac{3}{4} \\= ^{10}C_{2}\times \frac{1}{3^{3\times 4}}$

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