\(F=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2019\cdot2020}\)
Trả lời:
\(F=1.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+...+\frac{1}{2019}.\frac{1}{2020}\)
\(F=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(F=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(F=\frac{1}{1}-\frac{1}{2020}\)
\(F=\frac{2019}{2020}\)
\(F=1.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+...+\frac{1}{2019}.\frac{1}{2020}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(=1-\frac{1}{2020}=\frac{2020}{2020}-\frac{1}{2020}=\frac{2019}{2020}\)
Vậy \(F=\frac{2019}{2020}.\)
=\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+..................+\(\frac{1}{2019.2020}\)
=\(\frac{1}{1}\)-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+.........+\(\frac{1}{2019}\)-\(\frac{1}{2020}\)
=\(1-\frac{1}{2020}\)
=\(\frac{2019}{2020}\)
\(F=1.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+...+\frac{1}{2019}.\frac{1}{2020}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2019.2020}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(=1-\frac{1}{2020}=\frac{2019}{2020}\)
\(F=1\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{4}+...+\frac{1}{2019}\cdot\frac{1}{2020}\)
\(F=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2019\cdot2020}\)
\(F=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(F=1-\frac{1}{2020}=\frac{2019}{2020}\)
\(1\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{3}+\frac{1}{3}\times\frac{1}{4}+...+\frac{1}{2019}\times\frac{1}{2020}\)
= \(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{2019\times2020}\)
= \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
= \(\frac{1}{1}-\frac{1}{2020}\)
= \(\frac{2019}{2020}\)
