A ratio of 5^{th} term from beginning to 5^{th} term from end in binomial expansion of left( {

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7.Binomial Theorem

hard

A ratio of the $5^{th}$ term from the beginning to the $5^{th}$ term from the end in the binomial expansion of $\left( {{2^{1/3}} + \frac{1}{{2{{\left( 3 \right)}^{1/3}}}}} \right)^{10}$ is

$1:2{\left( 6 \right)^{\frac{1}{3}}}$

$1:4{\left( 16 \right)^{\frac{1}{3}}}$

$4{\left( {36} \right)^{\frac{1}{3}}}\,:\,1$

$2{\left( {36} \right)^{\frac{1}{3}}}\,:\,1$

(JEE MAIN-2019)

Solution

$\frac{{{5^{{\text{th}}}}{\text{ term from begining }}}}{{{5^{{\text{th}}}}{\text{ term from end }}}} = \frac{{10{{\text{C}}_4}{{\left( {\frac{1}{{2\left( {{3^{1/3}}} \right)}}} \right)}^4}{2^{6/3}}}}{{10{{\text{C}}_4}{{(2)}^{4/3}}{{\left( {\frac{1}{{2\left( {{3^{1/3}}} \right)}}} \right)}^6}}}$

$=\frac{2^{2} 2^{-2} 3^{-4 / 3}}{2^{4} 2^{(4 / 3)-6} 3^{-2}}=3^{2 / 3} \cdot 2^{8 / 3}$

$=4 \cdot(36)^{1 / 3}$

Standard 11

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Solution

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