Question

The greatest term in the expansion of $\sqrt{3}{\left[1+\left(\frac{1}{\sqrt{3}}\right)\right]}^{20}$ is

• A. $\frac{26840}{9}$
• B. $\frac{24840}{9}$
• C. $\frac{25840}{9}$
• D. None of these

Answer 1

The correct option is C

$\frac{25840}{9}$

The explanation for the correct option.

Find the greatest term in the expansion.

We know the greatest term in the expansion ${\left(\mathrm{x}+\mathrm{a}\right)}^{\mathrm{n}}$ is $\frac{{\mathrm{T}}_{\mathrm{r}+1}}{{\mathrm{T}}_{\mathrm{r}}}\ge 1$ i.e., $\frac{\mathrm{n}-\mathrm{r}+1}{\mathrm{r}}\left|\frac{\mathrm{a}}{\mathrm{n}}\right|\ge 1$.

The greatest term in the expansion $\sqrt{3}{\left[1+\left(\frac{1}{\sqrt{3}}\right)\right]}^{20}$ is $\frac{20-\mathrm{r}+1}{\mathrm{r}}\left|\frac{1}{\sqrt{3}}\right|\ge 1$.

$$\begin{gathered} \Rightarrow \frac{21-\mathrm{r}}{\mathrm{r}}\ge \sqrt{3} \\ \Rightarrow 21-\mathrm{r}\ge \sqrt{3}\mathrm{r} \\ \Rightarrow 21\ge \mathrm{r}\left(1+\sqrt{3}\right) \\ \Rightarrow \mathrm{r}\le \frac{21}{\left(1+\sqrt{3}\right)} \\ \Rightarrow \mathrm{r}\le 7.692 \\ \Rightarrow \mathrm{r}=7 \end{gathered}$$

The greatest term in the expansion is ${\mathrm{T}}_8=\sqrt{3}\times {}^{20}\mathrm{C}_7{\left(\frac{1}{\sqrt{3}}\right)}^7$

$$\begin{gathered} =\sqrt{3}\times {}^{20}\mathrm{C}_7\frac{1}{27\sqrt{3}} \\ =\frac{20\times 19\times 17\times 16\times 15\times 14}{1\times 2\times 3\times 4\times 5\times 6\times 7}\times \frac{1}{27} \\ =\frac{25840}{9} \end{gathered}$$

Hence, option (c) is the correct option.