यदि $$\left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots . .+\frac{1}{\alpha+1012}\right)-\left(\frac{1}{2 \cdot 1}+\frac{1}{4 \cdot 3}+\frac{1}{6 \cdot 5}+\ldots . .+\frac{1}{2024 \cdot 2023}\right)=\frac{1}{2024}$$ है, तो $$\alpha$$ बराबर हे __________.