Tìm số tự nhiên x biết 1/3 + 1/6 + 1/10 + ... + 1/x ( x + 1 ):2 = 2019/2021 A. 2019/2021 B. 2021 C. 2020 D. 2019
Giải thích
Trả lời:
\[\frac{1}{3} + \frac{1}{6} + \frac{1}{{10}} + ... + \frac{1}{{x\left( {x + 1} \right):2}} = \frac{{2019}}{{2021}}\]
\[2.\left[ {\frac{1}{{2.3}} + \frac{1}{{3.4}} + ... + \frac{1}{{x\left( {x + 1} \right)}}} \right] = \frac{{2019}}{{2021}}\]
\[2.\left( {\frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{x} - \frac{1}{{x - 1}}} \right) = \frac{{2019}}{{2021}}\]
\[2.\left( {\frac{1}{2} - \frac{1}{{x + 1}}} \right) = \frac{{2019}}{{2021}}\]
\[1 - \frac{2}{{x + 1}} = \frac{{2019}}{{2021}}\]
\[\frac{2}{{x + 1}} = 1 - \frac{{2019}}{{2021}}\]
\[\frac{2}{{x + 1}} = \frac{2}{{2021}}\]
\[x + 1 = 2021\]
\[x = 2020\]
Đáp án cần chọn là: C