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

अनुक्रम \(a_1, a_2, a_3, \ldots\) का विचार कीजिए जिसके लिए \(a _1=1, a _2=2\) हैं तथा \(a _{ n +2}=\frac{2}{ a _{ n +1}}+ a _{ n }, n =1\), \(2,3, \ldots\) हैं। यदि \(\left(\frac{a_1+\frac{1}{a_2}}{a_3}\right) \cdot\left(\frac{a_2+\frac{1}{a_3}}{a_4}\right) \cdot\left(\frac{a_3+\frac{1}{a_4}}{a_5}\right) \cdots \cdot\left(\frac{a_{30}+\frac{1}{a_{31}}}{a_{32}}\right)=2^a\left({ }^{61} C_{31}\right)\) है, तो \(\alpha\) बराबर है :

  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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