Question

The coefficient of $x^2$ in the binomial expansion of ${\left(\frac{1}{3}{x}^{1/2}+x^{-1/4}\right)}^{10}$ is:

• A. $\frac{70}{243}$
• B. $\frac{60}{423}$
• C. $\frac{50}{13}$
• D. none of these

Answer 1

Answer: (A)

The general term in the expansion of
${\left(\frac{1}{3}{x}^{1/2}+x^{-1/4}\right)}^{10}$ is
$T_{r+1}={}^{10}{C}_r{\left(\frac{1}{3}{x}^{1/2}\right)}^{10-r}{\left(x^{-1/4}\right)}^r$
$={}^{10}{C}_r\frac{1}{3^{10-r}}{x}^{5-r/2}{x}^{-r/4}$
$={}^{10}{C}_r\frac{1}{3^{10-r}}{x}^{5-3r/4}$
For coefficient of $x^2$, we take
$5-\frac{3r}{4}=2$
$\Rightarrow 3=\frac{3r}{4}$
$\Rightarrow r=4$
$\therefore$ Coefficient of $x^2={}^{10}{C}_4\frac{1}{3^{10-4}}$
$=\frac{210}{729}$
$=\frac{70}{243}$