Tìm số tự nhiên x biết [ frac{1}{3} + frac{1}{6} + frac{1}{{10}} + ... + frac{1}{{x left( {x + 1} right):2}} = frac{{2019}}{{2021}} ]

* Hướng dẫn giải

\[\frac{1}{3} + \frac{1}{6} + \frac{1}{{10}} + ... + \frac{1}{{x\left( {x + 1} \right):2}} = \frac{{2019}}{{2021}}\]

\[2\left[ {\frac{1}{{2.3}} + \frac{1}{{3.4}} + ... + \frac{1}{{x\left( {x + 1} \right)}}} \right] = \frac{{2019}}{{2021}}\]

\(2.\left( {\frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{x} - \frac{1}{{x + 1}}} \right) = \frac{{2019}}{{2021}}\)

\(2\left( {\frac{1}{2} - \frac{1}{{x + 1}}} \right) = \frac{{2019}}{{2021}}\)

\(1 - \frac{2}{{x + 2}} = \frac{{2019}}{{2021}}\)

\(\frac{2}{{x + 1}} = 1 - \frac{{2019}}{{2021}}\)

\(\begin{array}{l}\frac{2}{{x + 1}} = \frac{2}{{2021}}\\x + 1 = 2021\\x = 2020\end{array}\)

Đáp án cần chọn là: C