To solve the problem, we need to find the values of α and β based on the given probabilities of independent events \( A_1, A_2, A_3, \ldots, A_{1006} \).
### Step-by-Step Solution:
1. **Understanding the Events and Their Probabilities**:
Each event \( A_i \) has a probability given by:
\[
P(A_i) = \frac{1}{2i} \quad \text{for } i = 1, 2, \ldots, 1006
\]
2. **Finding the Probability of None of the Events Occurring**:
The probability that none of the events occur is given by:
\[
P(A_1^c \cap A_2^c \cap \ldots \cap A_{1006}^c) = P(A_1^c) \cdot P(A_2^c) \cdots P(A_{1006}^c)
\]
Where \( P(A_i^c) = 1 - P(A_i) = 1 - \frac{1}{2i} = \frac{2i - 1}{2i} \).
3. **Calculating the Product**:
Thus, we have:
\[
P(A_1^c) = \frac{1}{2}, \quad P(A_2^c) = \frac{3}{4}, \quad P(A_3^c = \frac{5}{6}, \ldots, P(A_{1006}^c) = \frac{2011}{2012}
\]
Therefore, the probability that none of the events occur is:
\[
P(A_1^c) \cdot P(A_2^c) \cdots P(A_{1006}^c = \prod_{i=1}^{1006} \frac{2i - 1}{2i}
\]
4. **Rewriting the Product**:
The product can be rewritten as:
\[
P(A_1^c) \cdot P(A_2^c) \cdots P(A_{1006}^c = \frac{1 \cdot 3 \cdot 5 \cdots \cdot 2011}{2 \cdot 4 \cdots \cdot 2012}
\]
The numerator is the product of the first 1006 odd numbers, and the denominator is \( 2^{1006} \cdot 1006! \).
5. **Final Expression**:
Thus, we can express the probability as:
\[
P(A_1^c \cap A_2^c \cap \ldots \cap A_{1006}^c = \frac{(1006)!}{2^{1006} \cdot (1006)!^2} = \frac{(1006)!}{2^{1006} \cdot (1006!)^2}
\]
6. **Comparing with the Given Probability**:
We are given that this probability equals:
\[
\frac{\alpha!}{2^\alpha (\beta!)^2}
\]
From the expressions, we can equate:
\[
\alpha = 2012 \quad \text{and} \quad \beta = 1006
\]
7. **Verifying Conditions**:
- **Condition 1**: \( \beta \) is of the form \( 4k + 2 \):
\[
1006 = 4 \cdot 251 + 2 \quad \text{(True)}
\]
- **Condition 2**: \( \alpha = 2\beta \):
\[
2012 = 2 \cdot 1006 \quad \text{(True)}
\]
- **Condition 3**: \( \beta \) is a composite number:
\[
1006 \text{ is composite (divisible by 2, 503)} \quad \text{(True)}
\]
- **Condition 4**: \( \alpha \) is of the form \( 4k \):
\[
2012 = 4 \cdot 503 \quad \text{(True)}
\]
### Conclusion:
All conditions are satisfied, thus confirming our values of \( \alpha \) and \( \beta \).
