Let the ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of (sqrt[4]2+frac1sqrt[4]3)^ n, in the increasing powers of frac1sqrt[4]3 be sqrt[4]6: 1. If the sixth term from the beginning is fracalphasqrt[4]3, then alpha is equal to _________.

Fifth term from beginning = ^nC_4( 2^1 / 4 )^n - 4( 3^ - 1 / 4 )^4 Fifth term from end = (n - 5 + 1)^th term from begin = ^nC_n - 4( 2^1 / 4 )^3( 3^ - 1 / 4 )^n - 4 Given ^nC_42^n - 4 / 4 . 3^ - 1 / ^nC_n - 32^4 / 4 . 3^ - ( n - 4 / 4 ) = 6^1 / 4 => 6^n - 8 / 4 = 6^1 / 4 => n - 8 / 4 = 1 / 4 => n = 9 T_6 = T_5 + 1 = ^9C_5( 2^1 / 4 )^4( 3^ - 1 / 4 )^5 = ^9C_5 . 2 / 3^1 / 4 . 3 = 84 / 3^1 / 4 = alpha / 3^1 / 4 => alpha = 84.

Let the ratio of the fifth term from the beginning to t... | null (Mathematics)

Fifth term from beginning

= {}^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}}

term from begin

= {}^n{C_{n - 4}}{\left( {{2^{{1 \over 4}}}} \right)^3}{\left( {{3^{{{ - 1} \over 4}}}} \right)^{n - 4}}

Given

{{{}^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}}}

\Rightarrow {6^{{{n - 8} \over 4}}} = {6^{{1 \over 4}}}

\Rightarrow {{n - 8} \over 4} = {1 \over 4} \Rightarrow n = 9

{T_6} = {T_{5 + 1}} = {}^9{C_5}{\left( {{2^{{1 \over 4}}}} \right)^4}{\left( {{3^{{{ - 1} \over 4}}}} \right)^5}

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

\Rightarrow \alpha = 84.

Explanation

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