Mathematics - Sequence & Series Question with Solution | TestHub

Given $a_{n+2}=\frac{2}{a_{n+1}}+a_n$

 $\Rightarrow a_{n+2}{a}_{n+1}-a_n{a}_{n+1}=2$

Now, $T_r=a_{r+1}{a}_r$ is an A.P. with common difference $2$

Now, $T_1=a_1{a}_2=2, T_2=a_2{a}_3=4, ...., T_r=2r$

So $\left(\frac{a_1+\frac{1}{a_2}}{a_3}\right)·\left(\frac{a_2+\frac{1}{a_3}}{a_4}\right)\cdots \left(\frac{a_{30}+\frac{1}{a_{31}}}{a_{32}}\right)=\prod _{i=1}^{30}\left(\frac{a_i+\frac{1}{a_{i+1}}}{a_{i+2}}\right)$

$=\prod _{i=1}^{30}\left(\frac{a_i{a}_{i+1}+1}{a_{i+1}{a}_{i+2}}\right)=\prod _{i=1}^{30}\frac{T_r+1}{T_{r+1}}$

$=\prod _{i=1}^{30}\left(\frac{2r+1}{2r+2}\right)=\frac{3·5·7\cdots 61}{4·6·8\cdots 62}=\frac{1·2·3·4·5·6·7\cdots 61·62}{2{\left(4·6·8\cdots 62\right)}^2}$

$=\frac{62!}{2{\left(4·6·8\cdots 62\right)}^2}=\frac{62!}{2^{61}·{\left(31!\right)}^2}=\frac{62\left(61!\right)}{2^{61}·31\left(31!\right)\left(30!\right)}$

$=2^{-60}{\cdot }^{61}{C}_{30}$

$=2^{-60}·{}^{61}C_{31}$

Now on comparing with $2^{\alpha }\left({}^{61}C_{31}\right)$

We get $\alpha =-60$