JEE Main 20249 Apr, Evening shiftMathematicsMediumSequence and Series179× since 2002If (1α+1+1α+2+…..+1α+1012)−(12⋅1+14⋅3+16⋅5+……+12024⋅2023)=12024\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 \ldots+\frac{1}{2024 \cdot 2023}\right)=\frac{1}{2024}(α+11+α+21+…..+α+10121)−(2⋅11+4⋅31+6⋅51+……+2024⋅20231)=20241, then α\alphaα is equal to ___________.Your answerType your answer. Integers and decimals are accepted.CheckSkip · show solution
