Mathematics - Sequence & Series Question with Solution | TestHub

$\begin{array}{rl}& \left(\frac{1}{\alpha +1}+\frac{1}{\alpha +2}+\dots +\frac{1}{\alpha +2012}\right)\\ & -\left\{\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\dots +\left(\frac{1}{2023}-\frac{1}{2024}\right)\right\}=\frac{1}{2024}\\ & \Rightarrow \left(\frac{1}{\alpha +1}+\frac{1}{\alpha +2}+\dots +\frac{1}{\alpha +2012}\right)\\ & -\{\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\dots +\frac{1}{2023}\\ & -\frac{1}{2024}-2\left(\frac{1}{2}+\frac{1}{4}+\dots +\frac{1}{2022}\right)\}=\frac{1}{2024}\\ & \end{array}$ $\begin{array}{rl}& \Rightarrow \left(\frac{1}{\alpha +1}+\frac{1}{\alpha +2}+\dots +\frac{1}{\alpha +2012}\right)\\ & -\left(\frac{1}{1}+\frac{1}{2}+\dots +\frac{1}{2023}\right)\\ & +\frac{1}{2024}+\left(\frac{1}{1}+\frac{1}{2}+\dots +\frac{1}{1011}\right)=\frac{1}{2024}\\ & \Rightarrow \frac{1}{\alpha +1}+\frac{1}{\alpha +2}+\dots +\frac{1}{\alpha +2012}\\ & =\frac{1}{1012}+\frac{1}{1013}+\dots +\frac{1}{2023}\\ & \Rightarrow \alpha =1011\end{array}$