Cho hàm số f(x) = ln((2021x)/(x+1)). Tính tổng S = f'(1) + f'(2) +....+f'(2020)?

Đáp ánC.

\(f\left( x \right) = \ln \frac{{2020x}}{{x + 1}} \Rightarrow f'\left( x \right) = \frac{1}{{x\left( {x + 1} \right)}} = \frac{1}{x} - \frac{1}{{x + 1}}\)

Khi đó: \(S = f'\left( 1 \right) + f'\left( 2 \right) + ... + f'\left( {2020} \right) = \sum\limits_{k = 1}^{2020} {\left( {\frac{1}{k} - \frac{1}{{k + 1}}} \right) = 1 - \frac{1}{{2021}} = \frac{{2020}}{{2021}}.} \)