Question

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

• A. $\frac{{}^{20}{C}_8}{27}$
• B. $\frac{{}^{20}{C}_7}{27}$
• C. $\frac{{}^{20}{C}_9}{27}$
• D. $\frac{{}^{20}{C}_6}{27}$

Answer 1

The correct option is B $\frac{{}^{20}{C}_7}{27}$

Let $(r+1{)}^{th}$ term be the greatest term in the expansion of ${(1+\frac{1}{\sqrt{3}})}^{20}.$
So, $\frac{T_{r+1}}{T_r}=\frac{20-r+1}{r}\left(\frac{1}{\sqrt{3}}\right)\ge 1$
$\Rightarrow 20-r+1\ge \sqrt{3}r$
$\Rightarrow 21\ge r(\sqrt{3}+1).$
$\Rightarrow r\le \frac{21}{\sqrt{3}+1}$
$\Rightarrow r\le 7.686$
So, the greatest term occurs when $r=7$.
Hence, the greatest term in $\sqrt{3}{(1+\frac{1}{\sqrt{3}})}^{20}$ is
$T_8=\sqrt{3}{}^{20}{C}_7{\left(\frac{1}{\sqrt{3}}\right)}^7=\frac{{}^{20}{C}_7}{27}$