Consider the sequence a _ 1 , a _ 2 , a _ 3 , such that a _ 1 = 1,...

Given ${a}_{n + 2} = \dfrac{2}{{a}_{n + 1}} + {a}_{n}$

$⇒ {a}_{n + 2}{a}_{n + 1} - {a}_{n}{a}_{n + 1} = 2$

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(\dfrac{{a}_{1} + \dfrac{1}{{a}_{2}}}{{a}_{3}}\right) \cdot \left(\dfrac{{a}_{2} + \dfrac{1}{{a}_{3}}}{{a}_{4}}\right) ⋯ \left(\dfrac{{a}_{30} + \dfrac{1}{{a}_{31}}}{{a}_{32}}\right) = \mathop{∏}_{i = 1}^{30}\left(\dfrac{{a}_{i} + \dfrac{1}{{a}_{i + 1}}}{{a}_{i + 2}}\right)$</p><p>$= \mathop{∏}_{i = 1}^{30}\left(\dfrac{{a}_{i}{a}_{i + 1} + 1}{{a}_{i + 1}{a}_{i + 2}}\right) = \mathop{∏}_{i = 1}^{30}\dfrac{{T}_{r} + 1}{{T}_{r + 1}}$</p><p>$= \mathop{∏}_{i = 1}^{30}\left(\dfrac{2r + 1}{2r + 2}\right) = \dfrac{3 \cdot 5 \cdot 7 ⋯ 61}{4 \cdot 6 \cdot 8 ⋯ 62} = \dfrac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 ⋯ 61 \cdot 62}{2{\left(4 \cdot 6 \cdot 8 ⋯ 62\right)}^{2}}$</p><p>$= \dfrac{62 !}{2{\left(4 \cdot 6 \cdot 8 ⋯ 62\right)}^{2}} = \dfrac{62 !}{{2}^{61} \cdot {\left(31 !\right)}^{2}} = \dfrac{62\left(61 !\right)}{{2}^{61} \cdot 31\left(31 !\right)\left(30 !\right)}$</p><p>$= {2}^{- 60}{⋅}^{61}{C}_{30}$</p><p>$= {2}^{- 60} \cdot C31 61$</p><p>Now on comparing with${2}^{\alpha}\left(C31 61\right)$</p><p>We get$\alpha = - 60