Ta có công thức tổng quát sau: \(1-\frac{1}{1+2+\cdots+n}\)
\(=1-\frac{1}{\frac{n\left(n+1\right)}{2}}\)
\(=1-\frac{2}{n\left(n+1\right)}\)
\(=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}\)
\(=\frac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)
Ta có: \(\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\cdot\ldots\cdot\left(1-\frac{1}{1+2+3+\cdots+2018}\right)\)
\(=\frac{\left(2+2\right)\left(2-1\right)}{2\cdot\left(2+1\right)}\cdot\frac{\left(3+2\right)\left(3-1\right)}{3\cdot\left(3+1\right)}\cdot\ldots\cdot\frac{\left(2018+2\right)\left(2018-1\right)}{2018\cdot\left(2018+1\right)}\)
\(=\frac{4\cdot1}{2\cdot3}\cdot\frac{5\cdot2}{3\cdot4}\cdot\ldots\cdot\frac{2020\cdot2017}{2018\cdot2019}\)
\(=\frac{4\cdot5\cdot\ldots\cdot2020}{3\cdot4\cdot\ldots\cdot2019}\cdot\frac{1\cdot2\cdot\ldots\cdot2017}{2\cdot3\cdot\ldots\cdot2018}=\frac{2020}{3}\cdot\frac{1}{2018}=\frac{1010}{1009\cdot3}=\frac{1010}{3027}\)
