Ta có công thức tổng quát sau:
\(1-\frac{2}{n\left(n+1\right)}=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}\)
\(=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)
Ta có:\(\frac23\cdot\frac56\cdot\frac{9}{10}\cdot\ldots\cdot\frac{4949}{4950}\)
= \(\left(1-\frac13\right)\left(1-\frac16\right)\cdot\ldots\cdot\left(1-\frac{1}{4950}\right)\)
\(=\left(1-\frac26\right)\left(1-\frac{2}{12}\right)\cdot\ldots\cdot\left(1-\frac{2}{9900}\right)\)
\(=\left(1-\frac{2}{2\cdot3}\right)\left(1-\frac{2}{3\cdot4}\right)\cdot...\cdot\left(1-\frac{2}{99\cdot100}\right)\)
\(=\frac{\left(2+2\right)\left(2-1\right)}{2\left(2+1\right)}\cdot\frac{\left(3+2\right)\left(3-1\right)}{3\left(3+1\right)}\cdot\ldots\cdot\frac{\left(99+2\right)\left(99-1\right)}{99\left(99+1\right)}\)
\(=\frac{4\cdot1}{2\cdot3}\cdot\frac{5\cdot2}{3\cdot4}\cdot\ldots\cdot\frac{101\cdot98}{99\cdot100}\)
\(=\frac{4\cdot5\cdot\ldots\cdot101}{3\cdot4\cdot\ldots\cdot100}\cdot\frac{1\cdot2\cdot\ldots\cdot98}{2\cdot3\cdot\ldots\cdot99}=\frac{101}{3}\cdot\frac{1}{99}=\frac{101}{297}\)
