JEE Mains · Maths · STD 11 - 7. binomial theoram
Let the ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of \(\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^{n}\), in the increasing powers of \(\frac{1}{\sqrt[4]{3}}\) be \(\sqrt[4]{6}: 1\). If the sixth term from the beginning is \(\frac{\alpha}{\sqrt[4]{3}}\), then \(\alpha\) is equal to\(.......\)
- A \(84\)
- B \(83\)
- C \(82\)
- D \(86\)
Answer & Solution
Correct Answer
(A) \(84\)
Step-by-step Solution
Detailed explanation
\(\frac{T_{5}}{T_{n-1}}=\frac{{ }^{n} C_{4}\left(2^{1 / 4}\right)^{n-4}\left(3^{-1 / 4}\right)^{4}}{\left(2^{1 / 4}\right)^{4}\left(3^{-1 / 4}\right)^{n-4}}=\frac{\sqrt[4]{6}}{1}\) \(\Rightarrow 2^{\frac{n-8}{4}} 3^{\frac{n-8}{4}}=6^{1 / 4}\) \(\Rightarrow 6^{n-3}=6\)…
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