Mathematics · 2024
JEE Main · 9 April 2024, Shift 2 · Q24
If (1/(α+1)+1/(α+2)+… …+1/(α+1012))-(1/(2 · 1)+1/(4 · 3)+1/(6 · 5)+… …+1/(2024 · 2023))=1/2024, then α is equal to ____.
If $\displaystyle \left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots \ldots+\frac{1}{\alpha+1012}\right)-\left(\frac{1}{2 \cdot 1}+\frac{1}{4 \cdot 3}+\frac{1}{6 \cdot 5}+\ldots \ldots+\frac{1}{2024 \cdot 2023}\right)=\frac{1}{2024}$, then $\displaystyle \alpha$ is equal to $\displaystyle \_\_\_\_$.
Official answer
From NTA’s final answer key for this paper.
1011
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JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.