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QJEEM22Ma247a: JEE Main 2022 Mathematics - Class 11 JEEM22Ma247a Solution: Let the ratio of the fifth term from the beginning to the fifth …

Let the ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of (24+134)n\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^{\mathrm{n}} , in the increasing powers of 134\frac{1}{\sqrt[4]{3}} be 64:1\sqrt[4]{6}: 1 . If the sixth term from the beginning is α34\frac{\alpha}{\sqrt[4]{3}} , then α\alpha is equal to _________.

Step-by-Step Verified Solution

Fifth term from beginning =nC4(214)n−4(3−14)4= {}^n{C_4}{\left( {{2^{{1 \over 4}}}} \right)^{n - 4}}{\left( {{3^{{{ - 1} \over 4}}}} \right)^4}

Fifth term from end =(n−5+1)th= {(n - 5 + 1)^{th}} term from begin =nCn−4(214)3(3−14)n−4= {}^n{C_{n - 4}}{\left( {{2^{{1 \over 4}}}} \right)^3}{\left( {{3^{{{ - 1} \over 4}}}} \right)^{n - 4}}

Given nC42n−44 . 3−1nCn−3244 . 3−(n−44)=614{{{}^n{C_4}{2^{{{n - 4} \over 4}}}\,.\,{3^{ - 1}}} \over {{}^n{C_{n - 3}}{2^{{4 \over 4}}}\,.\,{3^{ - \left( {{{n - 4} \over 4}} \right)}}}} = {6^{{1 \over 4}}}

⇒6n−84=614\Rightarrow {6^{{{n - 8} \over 4}}} = {6^{{1 \over 4}}}

⇒n−84=14⇒n=9\Rightarrow {{n - 8} \over 4} = {1 \over 4} \Rightarrow n = 9

T6=T5+1=9C5(214)4(3−14)5{T_6} = {T_{5 + 1}} = {}^9{C_5}{\left( {{2^{{1 \over 4}}}} \right)^4}{\left( {{3^{{{ - 1} \over 4}}}} \right)^5}

=9C5 . 2314 . 3=84314=α314= {{{}^9{C_5}\,.\,2} \over {{3^{{1 \over 4}}}\,.\,3}} = {{84} \over {{3^{{1 \over 4}}}}} = {\alpha \over {{3^{{1 \over 4}}}}}

⇒α=84.\Rightarrow \alpha = 84.

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