Đặt A =\(\frac{1}{99x97}+\frac{1}{97x95}+...+\frac{1}{3x1}\)
2A =\(\frac{2}{99x97}+\frac{2}{97x95}+...+\frac{2}{3x1}\)
2A=\(\frac{1}{97}-\frac{1}{99}+\frac{1}{95}-\frac{1}{97}+...+\frac{1}{1}-\frac{1}{3}\)
2A=1-\(\frac{1}{99}\)=\(\frac{98}{99}\)
=> A=\(\frac{49}{99}\)