Ta có: \(\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\cdots+\frac{2}{99\cdot100}\)
\(=2\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{99\cdot100}\right)\)
\(=2\left(1-\frac12+\frac12-\frac13+\cdots+\frac{1}{99}-\frac{1}{100}\right)\)
\(=2\left(1-\frac{1}{100}\right)=2\cdot\frac{99}{100}=\frac{99}{50}\)
