[tex] \frac{2}{1*3}+ \frac{2}{3*5}+..+ \frac{2}{2015*2017} = 2( \frac{1}{1*3}+ \frac{1}{3*5}+..+ \frac{1}{2015*2017}) = \\ \\ =2* \frac{1}{2}(\frac{3-1}{1*3} + \frac{5-3}{3*5}+ \frac{7-5}{5*7}+...+ \frac{2017-2015}{2015*2017}) = \\ \\ = (\frac{3}{1*3}- \frac{1}{1*3}) + (\frac{5}{3*5}- \frac{3}{3*5}) + (\frac{7}{5*7}- \frac{5}{5*7})+...+ \frac{2017}{2015*2017}- \frac{2015}{2015*2017} = \\ \\ = (1 - \frac{1}{3})+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{7})+....+\frac{1}{2015}- \frac{1}{2017}[/tex]
De aici se reduce tot inafara de primul si ultimul termen al sumei si avem:
[tex]=1- \frac{1}{2017} = \frac{2016}{2017} [/tex]
