Consider the sequence $a_{1}, a_{2}, a_{3}, \ldots$ such that $a_{1}=1, a_{2}=2$ and $a_{n+2}=\frac{2}{a_{n+1}}+a_{n}$ for $\mathrm{n}=1,2,3, \ldots .$ If $$\left(\frac{\mathrm{a}_{1}+\frac{1}{\mathrm{a}_{2}}}{\mathrm{a}_{3}}\right) \cdot\left(\frac{\mathrm{a}_{2}+\frac{1}{\mathrm{a}_{3}}}{\mathrm{a}_{4}}\right) \cdot\left(\frac{\mathrm{a}_{3}+\frac{1}{\mathrm{a}_{4}}}{\mathrm{a}_{5}}\right) \ldots\left(\frac{\mathrm{a}_{30}+\frac{1}{\mathrm{a}_{31}}}{\mathrm{a}_{32}}\right)=2^{\alpha}\left({ }^{61} \mathrm{C}_{31}\right)$$, then $\alpha$ is equal to :

<p>${a_{n + 2}} = {2 \over {{a_{n + 1}}}} + {a_n}$</p> <p>$\Rightarrow {a_n}{a_{n + 1}} + 1 = {a_{n + 1}}{a_{n + 2}} - 1$</p> <p>$\Rightarrow {a_{n + 2}}{a_{n + 1}} - {a_n}\,.\,{a_{n + 1}} = 2$</p> <p>For</p> <p>$$\matrix{ {n = 1} & {{a_3}{a_2} - {a_1}{a_2} = 2} \cr {n = 2} & {{a_4}{a_3} - {a_3}{a_2} = 2} \cr {n = 3} & {{a_5}{a_4} - {a_4}{a_3} = 2} \cr {} & {\matrix{ . \cr . \cr . \cr . \cr } } \cr {n = n} & {{{{a_{n + 2}}{a_{n + 1}} - {a_n}{a_{n + 1}} = 2} \over {{a_{n + 2}}{a_{n + 1}} = 2n + {a_1}{a_2}}}} \cr } $$</p> <p>Now,</p> <p>$${{({a_1}{a_2} + 1)} \over {{a_2}{a_3}}}\,.\,{{({a_2}{a_3} + 1)} \over {{a_3}{a_4}}}\,.\,{{({a_3}{a_4} + 1)} \over {{a_4}{a_5}}}\,.\,.....\,.\,{{({a_{30}}{a_{31}} + 1)} \over {{a_{31}}{a_{32}}}}$$</p> <p>$= {3 \over 4}\,.\,{5 \over 6}\,.\,{7 \over 8}\,.\,....\,.\,{{61} \over {62}}$</p> <p>$= {2^{ - 60}}\left( {{}^{61}{C_{31}}} \right)$</p>