JEE Mains · Maths · STD 11 - 8. sequence and series
If \(\left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+1012}\right) \) \( -\left(\frac{1}{2 \cdot 1}+\frac{1}{4 \cdot 3}+\frac{1}{6 \cdot 5}+\ldots+\frac{1}{2024 \cdot 2023}\right) \) \( =\frac{1}{2024}, \) then \(\alpha\) is equal to-
- A \(1367\)
- B \(1058\)
- C \(1056\)
- D \(1011\)
Answer & Solution
Correct Answer
(D) \(1011\)
Step-by-step Solution
Detailed explanation
\( \left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012}\right) \) - \( \left\{\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{2023}-\frac{1}{2024}\right)\right\}=\frac{1}{2024} \) \(\Rightarrow \)…
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