Question

[thuan truong]

tính C=\(1-\frac{2}{2.3}\)+ \(1-\frac{2}{3\cdot4}\)+ .......+ \(1-\frac{2}{2019\cdot2020}\)

Answer 1

[Lê Tài Bảo Châu]

\(C=1-\frac{2}{2.3}+1-\frac{2}{3.4}+...+1-\frac{2}{2019.2020}\)

\(=2018-2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\right)\)

\(=2018-2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\right)\)

\(=2018-2\left(\frac{1}{2}-\frac{1}{2020}\right)\)

\(=2018-2.\frac{1009}{2020}\)

\(=2018-\frac{1009}{1010}\)

\(=\frac{2037171}{1010}\)