If ( 1 + 1 + 1+ 2 + 1 + 1012 ) - ( 1 2 · 1 + 1 4 · 3 + 1 6 · 5 + 1 2024 · 2023 ) = 1 2024 , Then is equal to _______ . [2024]

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Solution

Q.
If $(\frac{1}{α + 1} + \frac{1}{α + 2} + \dots + \frac{1}{α + 1012}) - (\frac{1}{2 \cdot 1} + \frac{1}{4 \cdot 3} + \frac{1}{6 \cdot 5} + \dots + \frac{1}{2024 \cdot 2023}) = \frac{1}{2024},$ Then $α$ is equal to _______ . [2024]

Ans.
(1011)

Given $(\frac{1}{α + 1} + \frac{1}{α + 2} + \dots + \frac{1}{α + 1012}) - (\frac{1}{2 \cdot 1} + \frac{1}{4 \cdot 3} + \dots + \frac{1}{2024 \cdot 2023}) = \frac{1}{2024} ...(i)$

Now, $\sum_{r = 1}^{1012} \frac{1}{2 r (2 r - 1)} = \sum_{r = 1}^{1012} [\frac{1}{2 r - 1} - \frac{1}{2 r}]$

$= [(1 - \frac{1}{2}) + (\frac{1}{3} - \frac{1}{4}) + \dots + (\frac{1}{2023} - \frac{1}{2024})]$

$= (1 + \frac{1}{3} + \dots + \frac{1}{2023}) - (\frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2024})$

$= (1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{2023}) - \frac{1}{2} (1 + \frac{1}{2} + \dots + \frac{1}{1012}) - \frac{1}{2} (1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{1011})$

$\Rightarrow \frac{1}{1012} + \frac{1}{1013} + \dots + \frac{1}{2023} - \frac{1}{2024}$

$\Rightarrow \frac{1}{α + 1} + \frac{1}{α + 2} + \dots + \frac{1}{α + 1012} = \frac{1}{2024} - \sum_{r = 1}^{1012} \frac{1}{2 r (2 r - 1)} = \frac{1}{1012} + \dots + \frac{1}{2023}$

$\Rightarrow α + 1012 = 2023 \Rightarrow α = 1011$

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