JEE Mains · Maths · STD 11 - 8. sequence and series
Consider the sequence \(a_{1}, a_{2}, a_{3}, \ldots \ldots\) such that\(a _{1}=1, a _{2}=2 \text { and } a _{ n +2}=\frac{2}{ a _{ n +1}}+ a _{ n } \text { for } n =1,2,3, \ldots\) If \(\left(\frac{a_{1}+\frac{1}{a_{2}}}{a_{3}}\right) \cdot\left(\frac{a_{2}+\frac{1}{a_{3}}}{a_{4}}\right) \cdot\left(\frac{a_{3}+\frac{1}{a_{4}}}{a_{5}}\right) \cdots\left(\frac{a_{30}+\frac{1}{a_{31}}}{a_{32}}\right)=2^{\alpha}\left({ }^{61} C _{31}\right) .\) then \(\alpha\) is equal to.
- A \(-30\)
- B \(-31\)
- C \(-60\)
- D \(-61\)
Answer & Solution
Correct Answer
(C) \(-60\)
Step-by-step Solution
Detailed explanation
\(a_{a+2} a_{n+1}-a_{n+1} \cdot a_{a}=2\) Series will satisfy \(a_{1} a_{2}, a_{2} a_{3}, a_{3} a_{4}, a_{4} a_{5}\) \(\frac{1.2}{2.2} 2.3 \quad 2.4\) \(a_{n}+\frac{1}{a_{n+1}}=\frac{a_{n+2}-\frac{1}{a_{n+1}}}{a_{n+2}}\) \(=1-\frac{1}{a_{n+1} a_{n+2}}\) \(=1-\frac{1}{2(r+1)}\)…
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