If the ratio of the fifth term from the begin | vedclass

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Question

(B) In the expansion of $(a + b)^n$,the $r$-th term from the beginning is $T_r = {^nC_{r-1}} a^{n-r+1} b^{r-1}$.The $5$-th term from the beginning is

Vedclass · Accepted Answer

$60 \sqrt{3}$

Vedclass · Answer

(B) In the expansion of $(a + b)^n$,the $r$-th term from the beginning is $T_r = {^nC_{r-1}} a^{n-r+1} b^{r-1}$.The $5$-th term from the beginning is $T_5 = {^nC_4} (2^{1/4})^{n-4} (3^{-1/4})^4 = {^nC_4} 2^{(n-4)/4} 3^{-1}$.The $5$-th term from the end is the $(n-5+2) = (n-3)$-th term from the beginning,which is $T_{n-3} = {^nC_{n-4}} (2^{1/4})^4 (3^{-1/4})^{n-4} = {^nC_4} 2^1 3^{-(n-4)/4}$.Given the ratio $\frac{T_5}{T_{n-3}} = \frac{\sqrt{6}}{1} = 6^{1/2}$.Substituting the terms: $\frac{{^nC_4} 2^{(n-4)/4} 3^{-1}}{{^nC_4} 2^1 3^{-(n-4)/4}} = 2^{(n-4)/4 - 1} \cdot 3^{(n-4)/4 - 1} = (2 \cdot 3)^{(n-4)/4 - 1} = 6^{(n-4)/4 - 1}$.Equating the exponents: $\frac{n-4}{4} - 1 = \frac{1}{2}$ $\Rightarrow \frac{n-4}{4} = \frac{3}{2}$ $\Rightarrow n-4 = 6$ $\Rightarrow n = 10$.The third term from the beginning is $T_3 = {^{10}C_2} (2^{1/4})^{10-2} (3^{-1/4})^2 = 45 \cdot 2^2 \cdot 3^{-1/2} = \frac{45 \cdot 4}{\sqrt{3}} = \frac{180}{\sqrt{3}} = 60 \sqrt{3}$.

If the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of $(\sqrt[4]{2} + \frac{1}{\sqrt[4]{3}})^n$ is $\sqrt{6} : 1$,then the third term from the beginning is:

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