Question

(1-1/1.2)+(1-1/2.3)+(1/3.4)+...+(1-1/2022.2023)

Answer 1

Ta có: \(\left(1-\frac{1}{1\cdot2}\right)+\left(1-\frac{1}{2\cdot3}\right)+\cdots+\left(1-\frac{1}{2022\cdot2023}\right)\)

\(=2022-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{2022\cdot2023}\right)\)

\(=2022-\left(1-\frac12+\frac12-\frac13+\cdots+\frac{1}{2022}-\frac{1}{2023}\right)\)

\(=2022-\left(1-\frac{1}{2023}\right)=2022-\frac{2022}{2023}=2022\left(1-\frac{1}{2023}\right)\)

\(=2022\cdot\frac{2022}{2023}=\frac{4088484}{2023}\)