[tex]\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+\dfrac{2}{99}+...+\dfrac{2}{9999}= \\ \\ =\dfrac{2}{3\cdot 5}+\dfrac{2}{5\cdot 7}+\dfrac{2}{7\cdot 9}+\dfrac{2}{9\cdot11}+...+\dfrac{2}{99\cdot 101} = \\ \\ =\dfrac{5-3}{5\cdot 3}+\dfrac{7-5}{ 7\cdot 5}+\dfrac{9-7}{ 9\cdot 7}+\dfrac{11-9}{11\cdot 9}+...+\dfrac{101-99}{101\cdot 99}= [/tex]
[tex] =\Big(\dfrac{5}{5\cdot 3}-\dfrac{3}{5\cdot 3}\Big) + \Big(\dfrac{7}{7\cdot 5}-\dfrac{5}{7\cdot 5}\Big) +\Big(\dfrac{9}{9\cdot 7}-\dfrac{7}{9\cdot 7}\Big) +\\ \\ +\Big(\dfrac{11}{11\cdot 9}-\dfrac{9}{11\cdot 9}\Big) +...+\Big(\dfrac{101}{101\cdot 99}- \dfrac{99}{101\cdot 99}\Big)= \\ \\ \\ = \Big(\dfrac{1}{3}-\dfrac{1}{5}\Big)+\Big(\dfrac{1}{5}-\dfrac{1}{7}\Big)+\Big(\dfrac{1}{7}-\dfrac{1}{9}\Big)+\Big(\dfrac{1}{9}-\dfrac{1}{11}\Big)+...+\Big(\dfrac{1}{99}-\dfrac{1}{101}\Big) =[/tex]
[tex]
= \dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+...+\dfrac{1}{99}-\dfrac{1}{101} = \\ \\ =\dfrac{1}{3}-\dfrac{1}{101} = \dfrac{^{{\big{101}\big\slash}}1}{3}-\dfrac{^{{\big{3}\big\slash}}1}{101} = \dfrac{101}{101\cdot 3}-\dfrac{3}{101\cdot 3} = \dfrac{101-3}{101\cdot 3} = \dfrac{98}{303}[/tex]
