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

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