Greatest term in expansion of √ 3 {left( {1 + frac{1}{{√ 3 }}} right)^{20}} is

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7.Binomial Theorem

hard

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

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

$\frac{{24840}}{9}$

$\frac{{26840}}{9}$

None of these

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

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

Standard 11

Mathematics

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(JEE MAIN-2022)

If the ${(r + 1)^{th}}$ term in the expansion of ${\left( {\sqrt[3]{{\frac{a}{{\sqrt b }}}} + \sqrt {\frac{b}{{\sqrt[3]{a}}}} } \right)^{21}}$ has the same power of $a$ and $b$, then the value of $r$ is

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(JEE MAIN-2020)

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

Solution

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