If (1 +1 +1 +2 + +1 +1012 ) - (1 2 1 +1 4 3 +1 6 5 + +1 2024 2023… - Vidyadip

MCQ

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
$1056$
✓
$1011$

✓

Answer

Correct option: D.

$1011$

d
$ \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$

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