माना $(r + 1)$ वां पद महत्तम पद है, तब

${T_{r + 1}} = \sqrt 3 .{\,^{20}}{C_r}{\left( {\frac{1}{{\sqrt 3 }}} \right)^r}$तथा ${T_r} = \sqrt 3 .\,{\,^{20}}{C_{r – 1}}{\left( {\frac{1}{{\sqrt 3 }}} \right)^{r – 1}}$

अब $\frac{{{T_{r + 1}}}}{{{T_r}}} = \frac{{20 – r + 1}}{r}\left( {\frac{1}{{\sqrt 3 }}} \right)$

$\therefore \,\,\,\,{T_{r + 1}} \ge {T_r} \Rightarrow 20 – r + 1 \ge \sqrt 3 r$

$ \Rightarrow 21 \ge r(\sqrt 3  + 1)\,\, \Rightarrow r \le \frac{{21}}{{\sqrt 3  + 1}}$$ \Rightarrow \,\,r \le 7.686 \Rightarrow r = 7$

अत: महत्तम पद

${T_8} = \sqrt 3 {\,^{20}}{C_7}{\left( {\frac{1}{{\sqrt 3 }}} \right)^7} = \frac{{25840}}{9}$