QJEEM24Ma3b0a: JEE Main 2024 Mathematics - Class 11 JEEM24Ma3b0a Solution: If , then is equal to ___________.

$\begin{aligned}& \frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012}- \\& \left(\frac{1}{2 \times 21}+\frac{1}{4 \times 3}+\ldots+\frac{1}{2024} \cdot \frac{1}{2023}\right)=\frac{1}{2024} \\& \quad \sum_{r=1}^{1012} \frac{1}{2 r(2 r-1)}=\sum_{r=1}^{1012}\left(\frac{1}{2 r-1}-\frac{1}{2 r}\right) \\& \quad=\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{2023}-\frac{1}{2024}\right) \\& \quad=\left(1+\frac{1}{3}+\frac{1}{5}+\ldots .+\frac{1}{2023}\right) \\& \quad-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\ldots+\frac{1}{2024}\right) \\& \quad=\left(1+\frac{1}{3}+\ldots+\frac{1}{2023}\right)-\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{1012}\right)\end{aligned}$

$\begin{aligned}& =\left(1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{2023}\right)-\frac{1}{2}\left(1+\frac{1}{2}+\ldots+\frac{1}{1011}\right) \\& \frac{-1}{2}\left(1+\frac{1}{2}+\ldots+\frac{1}{1012}\right) \\& =\frac{1}{1012}+\frac{1}{1013}+\ldots+\frac{1}{2023}-\frac{1}{2024} \\& \Rightarrow \alpha+1012=2023 \\& \Rightarrow \alpha=1011\end{aligned}$