Đặt \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{97\cdot99}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\left(1-\frac{1}{99}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(\frac{1}{97}-\frac{1}{97}\right)=\left(\frac{99}{99}-\frac{1}{99}\right)+0+...+0=\frac{98}{99}\)