Key Concepts and Formulas

• Recurrence Relations: Understanding how to manipulate and simplify given recurrence relations.
• Telescoping Sums/Products: Recognizing patterns that allow for cancellation of intermediate terms.
• Binomial Coefficients: Knowledge of the definition and properties of binomial coefficients, particularly $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.
• Double Factorials: Understanding the notation $n!!$ (product of numbers with the same parity up to $n$).

Step-by-Step Solution

Step 1: Analyze the Recurrence Relation

We are given the recurrence relation $a_{n+2} = \frac{2}{a_{n+1}} + a_n$ with initial conditions $a_1 = 1$ and $a_2 = 2$. To simplify this relation, we multiply both sides by $a_{n+1}$: $a_{n+2} a_{n+1} = 2 + a_n a_{n+1}$ Rearranging the terms, we get: $a_{n+2} a_{n+1} - a_n a_{n+1} = 2$ This form reveals a constant difference between consecutive products of terms, which is a strong indicator of a telescoping pattern.

Step 2: Calculate the First Few Terms of the Sequence

We need the first few terms of the sequence to understand its behavior and to verify the recurrence relation. Given $a_1 = 1$ and $a_2 = 2$. For $n=1$: $a_3 = \frac{2}{a_2} + a_1 = \frac{2}{2} + 1 = 1 + 1 = 2$ For $n=2$: $a_4 = \frac{2}{a_3} + a_2 = \frac{2}{2} + 2 = 1 + 2 = 3$ For $n=3$: $a_5 = \frac{2}{a_4} + a_3 = \frac{2}{3} + 2 = \frac{2+6}{3} = \frac{8}{3}$

Step 3: Derive a General Formula for the Product $a_n a_{n+1}$

From Step 1, we have $a_{k+2} a_{k+1} - a_k a_{k+1} = 2$. Let's write this for $k = 1, 2, \ldots, m$: For $k=1$: $a_3 a_2 - a_1 a_2 = 2$ For $k=2$: $a_4 a_3 - a_2 a_3 = 2$ For $k=3$: $a_5 a_4 - a_3 a_4 = 2$ ... For $k=m$: $a_{m+2} a_{m+1} - a_m a_{m+1} = 2$

Summing these equations, we observe a telescoping sum: $(a_3 a_2 - a_1 a_2) + (a_4 a_3 - a_2 a_3) + \ldots + (a_{m+2} a_{m+1} - a_m a_{m+1}) = 2m$ The intermediate terms cancel out, leaving: $a_{m+2} a_{m+1} - a_1 a_2 = 2m$ Substitute the initial values $a_1 = 1$ and $a_2 = 2$, so $a_1 a_2 = 2$: $a_{m+2} a_{m+1} - 2 = 2m$ $a_{m+2} a_{m+1} = 2m + 2 = 2(m+1)$ Let $n = m+2$. Then $m = n-2$. Substituting this into the equation, we get the general formula for the product of consecutive terms: $a_n a_{n+1} = 2(n-1)$ This formula is valid for $n-2 \ge 1$, so $n \ge 3$. Let's check for $n=1$ and $n=2$. For $n=1$: $a_1 a_2 = 2(1-1) = 0$, which is incorrect as $a_1 a_2 = 1 \cdot 2 = 2$. For $n=2$: $a_2 a_3 = 2(2-1) = 2$, which is correct as $a_2 a_3 = 2 \cdot 2 = 4$. This indicates the formula $a_n a_{n+1} = 2(n-1)$ is valid for $n \ge 2$. Let's re-evaluate the sum.

The relation $a_{k+2}a_{k+1} - a_k a_{k+1} = 2$ can be rewritten as $(a_{k+2}-a_k)a_{k+1} = 2$. Let's use the form $a_{n+2}a_{n+1} - a_n a_{n+1} = 2$. For $n=1$: $a_3 a_2 - a_1 a_2 = 2$. This is $2 \cdot 2 - 1 \cdot 2 = 4 - 2 = 2$. Correct. For $n=2$: $a_4 a_3 - a_2 a_3 = 2$. This is $3 \cdot 2 - 2 \cdot 2 = 6 - 4 = 2$. Correct. So, $a_{n+2} a_{n+1} = a_n a_{n+1} + 2$. Let $b_n = a_n a_{n+1}$. Then $b_{n+1} = b_n + 2$. This is an arithmetic progression for $b_n$. $b_1 = a_1 a_2 = 1 \cdot 2 = 2$. $b_2 = a_2 a_3 = 2 \cdot 2 = 4$. $b_3 = a_3 a_4 = 2 \cdot 3 = 6$. The general term for $b_n$ is $b_n = b_1 + (n-1)d$, where $d=2$. $b_n = 2 + (n-1)2 = 2 + 2n - 2 = 2n$. So, $a_n a_{n+1} = 2n$. Let's verify this formula: For $n=1$: $a_1 a_2 = 2(1) = 2$. Correct. For $n=2$: $a_2 a_3 = 2(2) = 4$. Correct. For $n=3$: $a_3 a_4 = 2(3) = 6$. Correct. Thus, the correct relation is $a_n a_{n+1} = 2n$.

Step 4: Simplify the Given Product

The given product is: $P = \left(\frac{a_1 + \frac{1}{a_2}}{a_3}\right) \cdot \left(\frac{a_2 + \frac{1}{a_3}}{a_4}\right) \cdot \left(\frac{a_3 + \frac{1}{a_4}}{a_5}\right) \cdots \left(\frac{a_{30} + \frac{1}{a_{31}}}{a_{32}}\right)$ We can rewrite each term in the product by multiplying the numerator by $a_{n+1}$: $\frac{a_n + \frac{1}{a_{n+1}}}{a_{n+2}} = \frac{\frac{a_n a_{n+1} + 1}{a_{n+1}}}{a_{n+2}} = \frac{a_n a_{n+1} + 1}{a_{n+1} a_{n+2}}$ Now substitute the formula $a_n a_{n+1} = 2n$: $\frac{a_n a_{n+1} + 1}{a_{n+1} a_{n+2}} = \frac{2n + 1}{2(n+1)}$ The product becomes: $P = \prod_{n=1}^{30} \left(\frac{2n + 1}{2(n+1)}\right)$ Let's write out the terms of the product: $P = \left(\frac{2(1) + 1}{2(1+1)}\right) \cdot \left(\frac{2(2) + 1}{2(2+1)}\right) \cdot \left(\frac{2(3) + 1}{2(3+1)}\right) \cdots \left(\frac{2(30) + 1}{2(30+1)}\right)$ $P = \left(\frac{3}{4}\right) \cdot \left(\frac{5}{6}\right) \cdot \left(\frac{7}{8}\right) \cdots \left(\frac{61}{62}\right)$ This product can be written as: $P = \frac{3 \cdot 5 \cdot 7 \cdots 61}{4 \cdot 6 \cdot 8 \cdots 62}$

Step 5: Relate the Product to Binomial Coefficients

We need to express $P$ in the form $2^{\alpha} \binom{61}{31}$. The numerator is the product of odd numbers from 3 to 61. Let's include 1 for convenience: $1 \cdot 3 \cdot 5 \cdots 61$. The denominator is the product of even numbers from 4 to 62. Let's factor out 2 from each term: $2 \cdot 2 \cdot 3 \cdot 2 \cdot 4 \cdots 2 \cdot 31 = 2^{30} (2 \cdot 3 \cdot 4 \cdots 31) = 2^{30} \cdot 31!$.

Let's rewrite the numerator $1 \cdot 3 \cdot 5 \cdots 61$. We can express this using factorials. Consider $61! = (1 \cdot 3 \cdot 5 \cdots 61) \cdot (2 \cdot 4 \cdot 6 \cdots 60)$. The term $(2 \cdot 4 \cdot 6 \cdots 60) = 2^{30} (1 \cdot 2 \cdot 3 \cdots 30) = 2^{30} \cdot 30!$. So, $61! = (1 \cdot 3 \cdot 5 \cdots 61) \cdot 2^{30} \cdot 30!$. This gives $1 \cdot 3 \cdot 5 \cdots 61 = \frac{61!}{2^{30} \cdot 30!}$.

The product $P$ is $\frac{3 \cdot 5 \cdot 7 \cdots 61}{4 \cdot 6 \cdot 8 \cdots 62}$. Let's rewrite the numerator including 1: $1 \cdot 3 \cdot 5 \cdots 61 = \frac{61!}{2^{30} 30!}$. The denominator is $4 \cdot 6 \cdot 8 \cdots 62 = 2^{30} (2 \cdot 3 \cdot 4 \cdots 31) = 2^{30} \cdot 31!$. So, $P = \frac{61! / (2^{30} \cdot 30!)}{2^{30} \cdot 31!} = \frac{61!}{2^{30} \cdot 30! \cdot 2^{30} \cdot 31!} = \frac{61!}{2^{60} \cdot 30! \cdot 31!}$.

We know that $\binom{61}{31} = \frac{61!}{31! (61-31)!} = \frac{61!}{31! 30!}$. So, $P = \frac{1}{2^{60}} \cdot \frac{61!}{30! 31!} = \frac{1}{2^{60}} \binom{61}{31} = 2^{-60} \binom{61}{31}$.

Comparing this with the given form $2^{\alpha} \binom{61}{31}$, we have $\alpha = -60$.

Common Mistakes & Tips

• Incorrectly simplifying the recurrence relation: Ensure the product $a_n a_{n+1}$ is correctly derived. A common error is misinterpreting the sum or difference.
• Errors in factorial manipulation: Be careful when converting products of odd/even numbers into factorial forms. Double-check the powers of 2 and the factorial terms.
• Off-by-one errors in product limits: The product runs from $n=1$ to $n=30$. Ensure the terms in the final expression correspond to these limits.

Summary

The problem involves a sequence defined by a linear recurrence relation. By manipulating the recurrence relation, we were able to establish a formula for the product of consecutive terms, $a_n a_{n+1} = 2n$. Substituting this into the given product, we obtained a product of rational terms. This product was then simplified by relating it to factorials and subsequently to a binomial coefficient. By comparing the simplified product with the given expression, we determined the value of $\alpha$.

The final answer is $\boxed{-60}$.