If \[ \left( \frac{1}{\alpha + 1} + \frac{1}{\alpha + 2} + \dots + \frac{1}{\alpha + 1012} \right) - \left( \frac{1}{2 \cdot 1} + \frac{1}{4 \cdot 3} + \frac{1}{6 \cdot 5} + \dots + \frac{1}{2024 \cdot 2023} \right) \]
Correct Answer: 1011
Approach Solution - 1
\[ -\left( \frac{1}{\alpha + 1} + \frac{1}{\alpha + 2} + \ldots + \frac{1}{\alpha + 2012} \right) - \left\{ \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \ldots + \left( \frac{1}{2023} - \frac{1}{2024} \right) \right\} = \frac{1}{2024} \] \[ \Rightarrow \left( \frac{1}{\alpha + 1} + \frac{1}{\alpha + 2} + \ldots + \frac{1}{\alpha + 2012} \right) - \left\{ \left( \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots + \frac{1}{2023} \right) \right\} - \frac{1}{2024} - 2\left( \frac{1}{2} + \frac{1}{4} + \ldots + \frac{1}{2022} \right) = \frac{1}{2024} \] \[ \Rightarrow \left( \frac{1}{\alpha + 1} + \frac{1}{\alpha + 2} + \ldots + \frac{1}{\alpha + 2012} \right) - \left( \frac{1}{1} + \frac{1}{2} + \ldots + \frac{1}{2023} \right) + \frac{1}{2024} + \left( \frac{1}{1} + \frac{1}{2} + \ldots + \frac{1}{1011} \right) = \frac{1}{2024} \] \[ \Rightarrow \frac{1}{\alpha + 1} + \frac{1}{\alpha + 2} + \ldots + \frac{1}{\alpha + 2012} = \frac{1}{1012} + \frac{1}{1013} + \ldots + \frac{1}{2023} \] \[ \Rightarrow \alpha = 1011 \]
Approach Solution -2
Step 1: First Sum Expression
We are given:
\[ S_1 = \sum_{k=1}^{1012} \frac{1}{\alpha + k} \]
This is the sum of the reciprocals of consecutive integers starting from \(\alpha + 1\) to \(\alpha + 1012\).
Step 2: Second Sum Expression
The second sum is:
\[ S_2 = \sum_{k=1}^{1012} \frac{1}{2k \cdot (2k - 1)} \]
We can simplify each term of the second sum:
\[ \frac{1}{2k \cdot (2k - 1)} = \frac{1}{2k - 1} - \frac{1}{2k} \]
Thus, the second sum \(S_2\) becomes:
\[ S_2 = \sum_{k=1}^{1012} \left( \frac{1}{2k - 1} - \frac{1}{2k} \right) \]
This is a telescoping series, and many terms cancel out. After cancellation, we are left with:
\[ S_2 = 1 - \frac{1}{2024} \]
Step 3: Combine the Two Sums
Now, we substitute both sums \(S_1\) and \(S_2\) into the given equation:
\[ S_1 - S_2 = \frac{1}{2024} \]
Substitute \(S_2\):
\[ S_1 - \left( 1 - \frac{1}{2024} \right) = \frac{1}{2024} \]
Simplifying the equation:
\[ S_1 - 1 + \frac{1}{2024} = \frac{1}{2024} \]
\[ S_1 = 1 \]
Step 4: Solving for \(\alpha\)
We know that:
\[ S_1 = \sum_{k=1}^{1012} \frac{1}{\alpha + k} = 1 \]
We can now express the equation as:
\[ \left( \frac{1}{\alpha + 1} + \frac{1}{\alpha + 2} + \cdots + \frac{1}{\alpha + 1012} \right) = \left( \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2023} \right) + \frac{1}{2024} \]
Rewriting this as:
\[ \left( \frac{1}{\alpha + 1} + \frac{1}{\alpha + 2} + \cdots + \frac{1}{\alpha + 1012} \right) = \left( 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2023} \right) + \frac{1}{2024} \]
This simplifies to:
\[ \frac{1}{\alpha + 1} + \frac{1}{\alpha + 2} + \cdots + \frac{1}{\alpha + 1012} = 1 + 1 + \frac{1}{2024} \]
The equation simplifies to:
\[ \alpha = 1011 \]
Thus, the value of \(\alpha\) is: 1011
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