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Sequences and Series question

2024 · 9 Apr · Shift 2 · Q60
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Sequences and Series question

2024 · 9 Apr · Shift 2 · Q60

JEE MainMathematicsSequences and SeriesNumerical+4 / −1
If (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 ‾\underline{\hspace{2cm}}​.
Numerical answer
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