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 $$n = 1, 2, 3, \ldots$$. If $$\frac{a_1 + \frac{1}{a_2}}{a_3} \cdot \frac{a_2 + \frac{1}{a_3}}{a_4} \cdot \frac{a_3 + \frac{1}{a_4}}{a_5} \cdots \frac{a_{30} + \frac{1}{a_{31}}}{a_{32}} = 2^\alpha \cdot {}^{61}C_{31}$$ then $$\alpha$$ is equal to

We are given the sequence $$a_1 = 1, a_2 = 2$$ with recurrence $$a_{n+2} = \frac{2}{a_{n+1}} + a_n$$, and need to evaluate the product: $$\prod_{k=1}^{30} \frac{a_k + \frac{1}{a_{k+1}}}{a_{k+2}} = 2^\alpha \cdot \binom{61}{31}$$.

From the recurrence $$a_{n+2} = \frac{2}{a_{n+1}} + a_n$$ we obtain $$\frac{1}{a_{n+1}} = \frac{a_{n+2} - a_n}{2}$$. Substituting this into the numerator gives $$a_k + \frac{1}{a_{k+1}} = a_k + \frac{a_{k+2} - a_k}{2} = \frac{a_k + a_{k+2}}{2}$$, so each factor can be written as $$\frac{a_k + a_{k+2}}{2\cdot a_{k+2}}$$.

Next, multiplying numerator and denominator by $$a_{k+1}$$ yields

$$\frac{a_k + \frac{1}{a_{k+1}}}{a_{k+2}} = \frac{a_k \cdot a_{k+1} + 1}{a_{k+1} \cdot a_{k+2}}\,. $$

Introducing $$b_k = a_k \cdot a_{k+1}$$, observe that

$$b_{k+1} = a_{k+1} \cdot a_{k+2} = a_{k+1}\Bigl(\frac{2}{a_{k+1}} + a_k\Bigr) = 2 + a_k \cdot a_{k+1} = b_k + 2\,. $$

Since $$b_1 = a_1 \cdot a_2 = 1 \times 2 = 2$$, it follows by induction that $$b_k = 2k\,. $$ Therefore each factor simplifies to

$$\frac{b_k + 1}{b_{k+1}} = \frac{2k + 1}{2(k + 1)}\,. $$

Hence the desired product is

$$\prod_{k=1}^{30} \frac{2k + 1}{2(k + 1)} = \frac{1}{2^{30}} \cdot \prod_{k=1}^{30} \frac{2k + 1}{k + 1} = \frac{1}{2^{30}} \cdot \frac{3 \cdot 5 \cdot 7 \cdots 61}{2 \cdot 3 \cdot 4 \cdots 31}\,. $$

The numerator can be expressed as $$3 \cdot 5 \cdot 7 \cdots 61 = \frac{62!}{2^{31} \cdot 31!}\,, $$ while the denominator is $$2 \cdot 3 \cdot 4 \cdots 31 = 31!\,. $$ Hence the product becomes

$$\frac{1}{2^{30}} \cdot \frac{62!}{2^{31} \cdot 31! \cdot 31!} = \frac{62!}{2^{61} \cdot (31!)^2} = \frac{1}{2^{61}} \binom{62}{31}\,. $$

Since $$\binom{62}{31} = 2 \cdot \binom{61}{31}\,, $$ it follows that

$$\prod_{k=1}^{30} \frac{a_k + \frac{1}{a_{k+1}}}{a_{k+2}} = \frac{2 \cdot \binom{61}{31}}{2^{61}} = 2^{-60} \cdot \binom{61}{31}\,, $$

and therefore $$\alpha = -60\,. $$

The correct answer is Option C: $$-60\,. $$