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JEE Mains · Maths · STD 11 - 8. sequence and series

શ્રેણી \(a _{1}, a _{2}, a _{3}, \ldots .\) ધ્યાને લો કે જેથી \(a _{1}=1, a _{2}=2\) અને \(a _{ n +2}=\frac{2}{ a _{ n +1}}+ a _{ n }\) જ્યાં \(n =1,2,3, \ldots\). કે \(n =1,2,3, \ldots .\) If \(\left(\frac{ a _{1}+\frac{1}{ a _{2}}}{ a _{3}}\right) \cdot\left(\frac{ a _{2}+\frac{1}{ a _{3}}}{ a _{4}}\right) \cdot\left(\frac{ a _{3}+\frac{1}{ a _{4}}}{ a _{5}}\right) \ldots\left(\frac{ a _{30}+\frac{1}{ a _{31}}}{ a _{32}}\right)=2^{\alpha}\left({ }^{61} C _{31}\right)\), તો \(\alpha=\) .............

  1. A \(-30\)
  2. B \(-31\)
  3. C \(-60\)
  4. D \(-61\)
Verified Solution

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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