\(=2\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{100}\right)=2\left(\frac{97}{300}\right)=\frac{97}{150}\)
\(\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+...+\frac{4}{99\cdot101}=2\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\right)\)
\(=2+\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)=2\left[\left(\frac{1}{3}-\frac{1}{101}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+...+\left(\frac{1}{99}-\frac{1}{99}\right)\right]\)\(=2\left[\left(\frac{101}{303}-\frac{3}{303}\right)+0+...+0\right]=2\cdot\frac{98}{303}=\frac{196}{303}\)