Let the ratio of the fifth term from the begi | vedclass

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Question

(A) The binomial expansion is $(\sqrt[4]{2} + 3^{-1/4})^n$. Let $T_r$ denote the $r$-th term from the beginning. The $5$-th term from the beginning is

Vedclass · Accepted Answer

$84$

Vedclass · Answer

(A) The binomial expansion is $(\sqrt[4]{2} + 3^{-1/4})^n$. Let $T_r$ denote the $r$-th term from the beginning. The $5$-th term from the beginning is $T_5 = {^nC_4} (2^{1/4})^{n-4} (3^{-1/4})^4$. The $5$-th term from the end is the $(n-5+2) = (n-3)$-th term from the beginning,which is $T_{n-3} = {^nC_{n-4}} (2^{1/4})^4 (3^{-1/4})^{n-4} = {^nC_4} (2^{1/4})^4 (3^{-1/4})^{n-4}$. The ratio $\frac{T_5}{T_{n-3}} = \frac{(2^{1/4})^{n-4} (3^{-1/4})^4}{(2^{1/4})^4 (3^{-1/4})^{n-4}} = \frac{2^{(n-8)/4}}{3^{(8-n)/4}} = 2^{(n-8)/4} \cdot 3^{(n-8)/4} = 6^{(n-8)/4}$. Given $6^{(n-8)/4} = 6^{1/4}$,so $n-8 = 1$,which gives $n = 9$. The $6$-th term from the beginning is $T_6 = {^9C_5} (2^{1/4})^4 (3^{-1/4})^5 = 126 \cdot 2 \cdot 3^{-5/4} = \frac{252}{3 \cdot 3^{1/4}} = \frac{84}{3^{1/4}}$. Thus,$\alpha = 84$.

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