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 -

Question

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-

Answer 1

Correct answer is : 1011

\(\left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012}\right)\)

\( -\left\{\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{2023}-\frac{1}{2024}\right)\right\}=\frac{1}{2024} \)

\( \Rightarrow \left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012}\right) \)

\( -\left\{\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\ldots+\frac{1}{2023}\right. \) 

\( \left.-\frac{1}{2024}-2\left(\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2022}\right)\right\}=\frac{1}{2024} \)

\(\Rightarrow \left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012}\right) \)

\( -\left(\frac{1}{1}+\frac{1}{2}+\ldots+\frac{1}{2023}\right)\)

\(+\frac{1}{2024}+\left(\frac{1}{1}+\frac{1}{2}+\ldots+\frac{1}{1011}\right)=\frac{1}{2024} \)

\(\Rightarrow \frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012} \)

\(\frac{1}{1012}+\frac{1}{1013}+\ldots+\frac{1}{2023} \)

\(\Rightarrow \alpha=1011\)