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$\sqrt 3 {\left( {1 + \frac{1}{{\sqrt 3 }}} \right)^{20}}$ ના વિસ્તરણમાં મહતમ પદ મેળવો.
$\frac{{25840}}{9}$
$\frac{{24840}}{9}$
$\frac{{26840}}{9}$
એકપણ નહીં.
Solution
(a) Let $(r + 1)^{th}$ 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 $\frac{{{T_{r + 1}}}}{{{T_r}}} = \frac{{20 – r + 1}}{r}\left( {\frac{1}{{\sqrt 3 }}} \right)$
$\therefore \,\,\,\,{T_{r + 1}} \ge {T_r} \Rightarrow 20 – r + 1 \ge \sqrt 3 r$
$ \Rightarrow 21 \ge r(\sqrt 3 + 1)\,\, \Rightarrow r \le \frac{{21}}{{\sqrt 3 + 1}}$
$ \Rightarrow \,\,r \le 7.686 \Rightarrow r = 7$
Hence the greatest term is
${T_8} = \sqrt 3 {\,^{20}}{C_7}{\left( {\frac{1}{{\sqrt 3 }}} \right)^7} = \frac{{25840}}{9}$
