Hướng dẫn giải
Đáp án đúng là: D
\(\frac{1}{2} + \frac{1}{{2 \cdot 3}} + \frac{1}{{3 \cdot 4}} + ... + \frac{1}{{2021 \cdot 2022}}\)
= \[1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{{2021}} - \frac{1}{{2022}}\]
= \(1 + \left( { - \frac{1}{2} + \frac{1}{2}} \right) + \left( { - \frac{1}{3} + \frac{1}{3}} \right) + \left( { - \frac{1}{4} + \frac{1}{4}} \right) + ... + \left( { - \frac{1}{{2021}} + \frac{1}{{2021}}} \right) - \frac{1}{{2022}}\)
= \(1 + 0 + 0 + 0 + ... + 0 - \frac{1}{{2022}}\)
= \(1 - \frac{1}{{2022}}\)
= \(\frac{{2021}}{{2022}}\)