Question

If \[ \left( \frac{1}{\alpha + 1} + \frac{1}{\alpha + 2} + \dots + \frac{1}{\alpha + 1012} \right) - \left( \frac{1}{2 \cdot 1} + \frac{1}{4 \cdot 3} + \frac{1}{6 \cdot 5} + \dots + \frac{1}{2024 \cdot 2023} \right) \]

Answer 1

Correct Answer: 1011

\[ -\left( \frac{1}{\alpha + 1} + \frac{1}{\alpha + 2} + \ldots + \frac{1}{\alpha + 2012} \right) - \left\{ \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \ldots + \left( \frac{1}{2023} - \frac{1}{2024} \right) \right\} = \frac{1}{2024} \] \[ \Rightarrow \left( \frac{1}{\alpha + 1} + \frac{1}{\alpha + 2} + \ldots + \frac{1}{\alpha + 2012} \right) - \left\{ \left( \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots + \frac{1}{2023} \right) \right\} - \frac{1}{2024} - 2\left( \frac{1}{2} + \frac{1}{4} + \ldots + \frac{1}{2022} \right) = \frac{1}{2024} \] \[ \Rightarrow \left( \frac{1}{\alpha + 1} + \frac{1}{\alpha + 2} + \ldots + \frac{1}{\alpha + 2012} \right) - \left( \frac{1}{1} + \frac{1}{2} + \ldots + \frac{1}{2023} \right) + \frac{1}{2024} + \left( \frac{1}{1} + \frac{1}{2} + \ldots + \frac{1}{1011} \right) = \frac{1}{2024} \] \[ \Rightarrow \frac{1}{\alpha + 1} + \frac{1}{\alpha + 2} + \ldots + \frac{1}{\alpha + 2012} = \frac{1}{1012} + \frac{1}{1013} + \ldots + \frac{1}{2023} \] \[ \Rightarrow \alpha = 1011 \] 

Answer 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