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
$T_{r+1}={ }^{10} C_{r}\left(\frac{1}{3} x^{1 / 2}\right)^{10-r}\left(x^{-1 / 4}\right)^{r}$
$={ }^{10} C_{r} \frac{1}{3^{10-r}} x^{5-r / 2} x^{-r / 4}$
$={ }^{10} C_{r} \frac{1}{3^{10-r}} x^{5-3 r / 4}$
For coefficient of $x^{2}$, we take
$5-\frac{3 r}{4}=2$
$\Rightarrow 3=\frac{3 r}{4} $
$ \Rightarrow r=4$
$\therefore $ Coefficient of $x^{2}={ }^{10} C_{4} \frac{1}{3^{10-4}}$
$=\frac{210}{729}$
$=\frac{70}{243}$