Cho hàm số f(x)=ln((x+1)/x). Tính tổng s=f’(1)+f’(2)+……+f’(2019). Bài này làm sao ạ 05/08/2021 Bởi Isabelle Cho hàm số f(x)=ln((x+1)/x). Tính tổng s=f’(1)+f’(2)+……+f’(2019). Bài này làm sao ạ
\(\begin{array}{l}f\left( x \right) = \ln \left( {\frac{{x + 1}}{x}} \right) = \ln \left( {x + 1} \right) – \ln x\\ \Rightarrow f’\left( x \right) = \frac{1}{{x + 1}} – \frac{1}{x}\end{array}\) Khi đó ta có: \(\begin{array}{l}S = f’\left( 1 \right) + f’\left( 2 \right) + … + f’\left( {2019} \right)\\S = \frac{1}{2} – \frac{1}{1} + \frac{1}{3} – \frac{1}{2} + \frac{1}{4} – \frac{1}{3} + … + \frac{1}{{2020}} – \frac{1}{{2019}}\\S = \frac{1}{{2020}} – 1 = – \frac{{2019}}{{2020}}\end{array}\) Bình luận
\(\begin{array}{l}f\left( x \right) = \ln \left( {\frac{{x + 1}}{x}} \right) = \ln \left( {x + 1} \right) – \ln x\\ \Rightarrow f’\left( x \right) = \frac{1}{{x + 1}} – \frac{1}{x}\end{array}\)
Khi đó ta có:
\(\begin{array}{l}S = f’\left( 1 \right) + f’\left( 2 \right) + … + f’\left( {2019} \right)\\S = \frac{1}{2} – \frac{1}{1} + \frac{1}{3} – \frac{1}{2} + \frac{1}{4} – \frac{1}{3} + … + \frac{1}{{2020}} – \frac{1}{{2019}}\\S = \frac{1}{{2020}} – 1 = – \frac{{2019}}{{2020}}\end{array}\)