If $\left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots \ldots+\frac{1}{\alpha+1012}\right)-\left(\frac{1}{2 \cdot 1}+\frac{1}{4 \cdot 3}+\frac{1}{6 \cdot 5}+\ldots . .+\frac{1}{2024 \cdot 2023}\right)=\frac{1}{2024}$, then $\alpha$ is equal to________

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Aicharya · Accepted Answer

$\begin{aligned} & \left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012}ight) \ & -\left\{\left(\frac{1}{1}-\frac{1}{2}ight)+\left(\frac{1}{3}-\frac{1}{4}ight)+\ldots+\left(\frac{1}{2023}-\frac{1}{2024}ight)ight\}=\frac{1}{2024} \ & \Rightarrow\left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012}ight) \ & -\left\{\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}ight)+\ldots+\frac{1}{2023}ight. \ & \left.-\frac{1}{2024}-2\left(\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2022}ight)ight\}=\frac{1}{2024} \ & \end{aligned}$ $\begin{aligned} & \Rightarrow\left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012}ight) \ & \quad-\left(\frac{1}{1}+\frac{1}{2}+\ldots+\frac{1}{2023}ight) \ & \quad+\frac{1}{2024}+\left(\frac{1}{1}+\frac{1}{2}+\ldots+\frac{1}{1011}ight)=\frac{1}{2024} \ & \Rightarrow \frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012} \ & \quad=\frac{1}{1012}+\frac{1}{1013}+\ldots+\frac{1}{2023} \ & \Rightarrow \alpha=1011\end{aligned}$

If $\left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots \ldots+\frac{1}{\alpha+1012}\right)-\left(\frac{1}{2…

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