Question

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

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

Answer 1

The correct option is A $\frac{25840}{9}$

Let (r+1) term be the greatest term. Then
$T_{r+1}=\sqrt{3}{.}^{20}{C}_r{\left(\frac{1}{\sqrt{3}}\right)}^r$ and $T_r=\sqrt{3}{.}^{20}{C}_{r-1}{\left(\frac{1}{\sqrt{3}}\right)}^{r-1}$
Now, $\frac{T_{r+1}}{T_r}=\frac{20-r+1}{r}\left(\frac{1}{\sqrt{3}}\right)$
$\therefore T_{r+1}\ge T_r\Rightarrow 20-r\ge \sqrt{3}r$
$\Rightarrow 21\ge r(\sqrt{3}+1)\Rightarrow \le \frac{21}{\sqrt{3}+1}\Rightarrow r\le 7.686\Rightarrow r=7$
Hence the greatest term is
$T_8=\sqrt{3}{}^{20}{C}_7{\left(\frac{1}{\sqrt{3}}\right)}^7=\frac{25840}{9}$