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(C) Given $a_{n+2} = \frac{2}{a_{n+1}} + a_{n}$,we have $a_{n+2} a_{n+1} = 2 + a_{n} a_{n+1}$.This implies $a_{n+2} a_{n+1} - a_{n+1} a_{n} = 2$.Let $

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$-60$

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(C) Given $a_{n+2} = \frac{2}{a_{n+1}} + a_{n}$,we have $a_{n+2} a_{n+1} = 2 + a_{n} a_{n+1}$.This implies $a_{n+2} a_{n+1} - a_{n+1} a_{n} = 2$.Let $b_{n} = a_{n} a_{n+1}$. Then $b_{n+1} - b_{n} = 2$,which is an arithmetic progression with $b_{1} = a_{1} a_{2} = 1 \cdot 2 = 2$ and common difference $d = 2$.Thus,$b_{n} = b_{1} + (n-1)d = 2 + (n-1)2 = 2n$.Now,consider the term $\frac{a_{n} + \frac{1}{a_{n+1}}}{a_{n+2}} = \frac{a_{n} a_{n+1} + 1}{a_{n+1} a_{n+2}} = \frac{b_{n} + 1}{b_{n+1}} = \frac{2n + 1}{2(n+1)}$.The product is $P = \prod_{n=1}^{30} \frac{2n+1}{2(n+1)} = \frac{3 \cdot 5 \cdot 7 \cdots 61}{2^{30} \cdot (2 \cdot 3 \cdots 31)}$.Multiply numerator and denominator by $2^{30} \cdot 30! = 2^{30} \cdot (1 \cdot 2 \cdots 30)$:$P = \frac{(1 \cdot 2 \cdot 3 \cdots 61) / (2^{30} \cdot 30!)}{2^{30} \cdot 31!} = \frac{61!}{2^{60} \cdot 31! \cdot 30!} = \frac{1}{2^{60}} \cdot \frac{61!}{31! \cdot 30!} = 2^{-60} \cdot {}^{61}C_{31}$.Comparing with $2^{\alpha} \cdot {}^{61}C_{31}$,we get $\alpha = -60$.

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