Question

Consider the sequence $a_1,a_2,a_3,\dots \dots$ such that $a_1=1,a_2=2$ and $a_{n+2}=\frac{2}{a_{n+1}}+a_n$ for $n=1,2,3,\dots$. If $\left(\frac{a_1+\frac{1}{a_2}}{a_3}\right)\cdot \left(\frac{a_2+\frac{1}{a_3}}{a_4}\right)\cdot \left(\frac{a_3+\frac{1}{a_4}}{a_5}\right)\cdot \cdots \cdot \left(\frac{a_{30}+\frac{1}{a_{31}}}{a_{32}}\right)=2^a\left({}^{61}{C}_{31}\right)$, then $\alpha$ is equal to :

• A. $-30$
• B. $-31$
• C. $-60$
• D. $-61$

Answer 1

Answer: (C)

$a_{n+2}{a}_{n+1}-a_{n+1}\cdot a_n=2$
Series will satisfy

$\frac{a_n+\frac{1}{a_{n+1}}}{a_{n+2}}=\frac{a_{n+2}-\frac{1}{a_{n+1}}}{a_{n+2}}$
$=1-\frac{1}{a_{n+1}{a}_{n+2}}$
$=1-\frac{1}{2(r+1)}$
$=\frac{2r+1}{2(r+1)}$
Now proof is given by
$=\prod _{r=1}^{30}\frac{(2r+1)}{2(r+1)}$
$=\frac{(1\cdot 3\cdot 5\cdot \dots \dots ..61)}{2^{30}\cdot (2\cdot 3\cdot \dots \dots .31)}$