Ta có: \(2y+\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{2020\cdot2021}\right)=\dfrac{4041}{2021}\)
\(\Leftrightarrow2y+\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2020}-\dfrac{1}{2021}\right)=\dfrac{4041}{2021}\)
\(\Leftrightarrow2y+1-\dfrac{1}{2021}=\dfrac{4041}{2021}\)
\(\Leftrightarrow2y=\dfrac{4041}{2021}+\dfrac{1}{2021}-1\)
\(\Leftrightarrow2y=2-1=1\)
hay \(y=\dfrac{1}{2}\)
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