$$\begin{gathered} Consider the sequence a_1, a_2, a_3, ........ such that \\ {a}_1=1, a_2=2, and a_{n+2}=\frac{2}{a_{n+1}}+a_n for n= 1, 2, 3, ....... \\ if \left[\frac{a_1+{\displaystyle \frac{1}{a_2}}}{a_3}\right]·\left[\frac{a_2+\frac{1}{a_3}}{a_4}\right]·\left[\frac{a_3+{\displaystyle \frac{1}{a_4}}}{a_5}\right]....\left[\frac{a_{30}+{\displaystyle \frac{1}{a_{31}}}}{a_{32}}\right]=2^{\alpha }\left({{61}_C}_{31}\right), \\ then \alpha is equal to : \end{gathered}$$

$$\begin{gathered} a_{n+2}{a}_{n+1}-a_{n+1}.a_n=2 \\ Series will satisfy \\ {a}_1{a}_2, a_2{a}_3, a_3{a}_4, a_4{a}_5, \\ 1.2 2.2 2.3 2.4 \\ \frac{a_n+{\displaystyle \frac{1}{a_{n+1}}}}{a_{n+2}}=\frac{a_{n+2}-{\displaystyle \frac{1}{a_{n+1}}}}{a_{n+2}} \\ =1-\frac{1}{a_{n+1}{a}_{n+2}} \\ =1-\frac{1}{2\left(r+1\right)} \\ =\frac{2r+1}{2\left(r+1\right)} \\ \mathrm{Now} \mathrm{product} \mathrm{is} \mathrm{given} \mathrm{by} \\ =\prod _{r=1}^{30}\frac{\left(2r+1\right)}{2\left(r+1\right)}=\frac{\left(1.3.5·........·61\right)}{2^{30}.\left(2·3·.....·31\right)} \\ =\frac{\left(1·3·5·.......·61\right)}{31.2^{30}!}\times \frac{2^{30}\times 30!}{2^{30}\times 30!} \\ =\frac{61!}{2^{60}31!·30!} \\ \therefore \alpha =-60 \end{gathered}$$