The greatest term in the expansion of sqrt 3 | vedclass

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Question

(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} =

Vedclass · Accepted Answer

$\frac{{25840}}{9}$

Vedclass · Answer

(a) Let $(r + 1)^{th}$ term be the greatest term.

Then ${T_{r + 1}} = \sqrt 3 .{\,^{20}}{C_r}{\left( {\frac{1}{{\sqrt 3 }}} ight)^r}$and ${T_r} = \sqrt 3 .\,{\,^{20}}{C_{r - 1}}{\left( {\frac{1}{{\sqrt 3 }}} ight)^{r - 1}}$ 

Now $\frac{{{T_{r + 1}}}}{{{T_r}}} = \frac{{20 - r + 1}}{r}\left( {\frac{1}{{\sqrt 3 }}} ight)$ 

$	herefore \,\,\,\,{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 

${T_8} = \sqrt 3 {\,^{20}}{C_7}{\left( {\frac{1}{{\sqrt 3 }}} ight)^7} = \frac{{25840}}{9}$

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

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