In the binomial (2^{1/3} + 3^{-1/3})^n,if the | vedclass

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(B) The general term of the expansion $(a + b)^n$ is given by $T_{r + 1} = ^nC_r a^{n - r} b^r$.Here,$a = 2^{1/3}$ and $b = 3^{-1/3}$.The $7^{th}$ ter

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$9$

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(B) The general term of the expansion $(a + b)^n$ is given by $T_{r + 1} = ^nC_r a^{n - r} b^r$.Here,$a = 2^{1/3}$ and $b = 3^{-1/3}$.The $7^{th}$ term from the beginning is $T_7 = ^nC_6 a^{n - 6} b^6$.The $7^{th}$ term from the end is the $7^{th}$ term from the beginning of the expansion $(b + a)^n$,which is $T_7' = ^nC_6 b^{n - 6} a^6$.The ratio is given as $\frac{T_7}{T_7'} = \frac{^nC_6 a^{n - 6} b^6}{^nC_6 b^{n - 6} a^6} = \frac{a^{n - 12}}{b^{n - 12}} = \left(\frac{a}{b}\right)^{n - 12} = \frac{1}{6}$.Substituting $a = 2^{1/3}$ and $b = 3^{-1/3}$,we get $\left(\frac{2^{1/3}}{3^{-1/3}}\right)^{n - 12} = (2^{1/3} \cdot 3^{1/3})^{n - 12} = (6^{1/3})^{n - 12} = 6^{\frac{n - 12}{3}}$.Given $6^{\frac{n - 12}{3}} = 6^{-1}$,we equate the exponents: $\frac{n - 12}{3} = -1$.$n - 12 = -3$,which gives $n = 9$.

In the binomial $(2^{1/3} + 3^{-1/3})^n$,if the ratio of the seventh term from the beginning of the expansion to the seventh term from its end is $1/6$,then $n =$

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