Find n, if the ratio of the fifth term from t | vedclass

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(A) Let the expansion be $(a + b)^n$. The $r$-th term from the beginning is $T_r = {\,^nC_{r-1}} a^{n-r+1} b^{r-1}$.The fifth term from the beginning

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(A) Let the expansion be $(a + b)^n$. The $r$-th term from the beginning is $T_r = {\,^nC_{r-1}} a^{n-r+1} b^{r-1}$.The fifth term from the beginning is $T_5 = {\,^nC_4} (2^{1/4})^{n-4} (3^{-1/4})^4 = {\,^nC_4} (2^{1/4})^{n-4} (3^{-1})$.The fifth term from the end is the $(n-5+1) = (n-4)$-th term from the beginning,which is $T_{n-4} = {\,^nC_{n-5}} (2^{1/4})^{n-(n-4)+1} (3^{-1/4})^{n-5} = {\,^nC_{n-5}} (2^{1/4})^5 (3^{-1/4})^{n-5}$.Alternatively,the fifth term from the end in $(a+b)^n$ is the fifth term from the beginning in $(b+a)^n$,which is $T'_5 = {\,^nC_4} (3^{-1/4})^{n-4} (2^{1/4})^4 = {\,^nC_4} (3^{-1/4})^{n-4} (2^1)$.Given the ratio $\frac{T_5}{T'_5} = \sqrt{6} : 1$,we have:$\frac{{\,^nC_4} (2^{1/4})^{n-4} (3^{-1})}{ {\,^nC_4} (3^{-1/4})^{n-4} (2^1) } = \sqrt{6}$.$\frac{2^{(n-4)/4}}{2^1} \cdot \frac{3^{(n-4)/4}}{3^1} = 6^{1/2}$.$2^{(n-4)/4 - 1} \cdot 3^{(n-4)/4 - 1} = 6^{1/2}$.$(2 \cdot 3)^{(n-4)/4 - 1} = 6^{1/2}$.$6^{(n-4-4)/4} = 6^{1/2}$.$\frac{n-8}{4} = \frac{1}{2}$ $\Rightarrow n-8 = 2$ $\Rightarrow n = 10$.

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

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