Consider the sequence a_ 1 , a_ 2 , a_ 3 , such thata _ 1 =1, a _… - Vidyadip

MCQ

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
$-31$
✓
$-60$
D
$-61$

✓

Answer

Correct option: C.

$-60$

c
$a_{a+2} a_{n+1}-a_{n+1} \cdot a_{a}=2$

Series will satisfy

$a_{1} a_{2}, a_{2} a_{3}, a_{3} a_{4}, a_{4} a_{5}$

$\frac{1.2}{2.2} 2.3 \quad 2.4$

$a_{n}+\frac{1}{a_{n+1}}=\frac{a_{n+2}-\frac{1}{a_{n+1}}}{a_{n+2}}$

$=1-\frac{1}{a_{n+1} a_{n+2}}$

$=1-\frac{1}{2(r+1)}$

$=\frac{2 r+1}{2(r+1)}$

Now proof is given by

$=\prod_{r=1}^{30} \frac{(2 r+1)}{2(r+1)}$

$=\frac{(1 \cdot 3 \cdot 5 \cdot \ldots \ldots \cdot 61)}{2^{30} \cdot(2 \cdot 3 \cdot \ldots \ldots \cdot 31)}$

$\Rightarrow \frac{(1 \cdot 3 \cdot 5 \cdot \ldots \ldots \ldots \cdot 61)}{\mid 31 \cdot 2^{30}} \times \frac{2^{30} \times \underline{30}}{2^{30} \times \underline{30}}$

$=\frac{\lfloor 61}{2^{60}|31 \cdot| 30}$

$\alpha=-60$

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