Question

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

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

Answer 1

The correct option is A

$\frac{70}{243}$

Expatiation for the correct option:

Given, the expression is ${\left[\frac{1}{3}{x}^{\frac{1}{2}}+{\left(x\right)}^{\frac{-1}{4}}\right]}^{10}$

Using Binomial expansion as the general term

$$\begin{gathered} T_{r+1}={}^{10}C_r{\left(\frac{1}{3}{x}^{\frac{1}{2}}\right)}^{10-r}{\left(x^{\frac{-1}{4}}\right)}^r \\ {T}_{r+1}={}^{10}C_r{\left(\frac{1}{3}\right)}^{10-r}{\left(x\right)}^{5-\frac{r}{2}-\frac{r}{4}} \end{gathered}$$

Now , in order to find coefficient of ${\mathrm{x}}^2$ comparing it we get,

$$\begin{gathered} 5-\frac{r}{2}-\frac{r}{4}=2 \\ r=4 \end{gathered}$$

So, coefficient of ${\mathrm{x}}^2$ will be ${}^{10}C_r{\left(\frac{1}{3}\right)}^{10-r}$ put $r=4$ we get

$$\begin{gathered} {}^{10}C_4{\left(\frac{1}{3}\right)}^{10-4} \\ \frac{70}{243} \end{gathered}$$

Hence, option (A) is correct.