Bài giải:

$\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}+\frac{1}{101.103}$

$=\frac{1}{2}.\left( \frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}+\frac{2}{101.103} \right)$

$=\frac{1}{2}.\left( 1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}+\frac{1}{101}-\frac{1}{103} \right)$

$=\frac{1}{2}.\left( 1-\frac{1}{103} \right)$

$=\frac{1}{2}.\frac{102}{103}$

$=\frac{51}{103}$