\(\frac{3}{3\cdot5}+\frac{3}{5\cdot7}+...+\frac{3}{97\cdot99}=2\cdot\frac{1}{2}\cdot\left(\frac{3}{3\cdot5}+\frac{3}{5\cdot7}+...+\frac{3}{97\cdot99}\right)\)
\(=3\cdot\frac{1}{2}\cdot\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\right)=\frac{3}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)\(=\frac{3}{2}\cdot\left[\left(\frac{1}{3}-\frac{1}{99}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{97}-\frac{1}{97}\right)\right]\)
\(=\frac{3}{2}\cdot\left[\left(\frac{33}{99}-\frac{1}{99}\right)+0+...+0\right]=\frac{3}{2}\cdot\frac{32}{99}=\frac{3\cdot32}{2\cdot99}=\frac{1\cdot16}{1\cdot33}=\frac{16}{33}\)
Chúc bạn học tốt !^_^
