Question

Dau coroana! 3/1*5+4 /5*9+4/9*13+.....+4/2004*2008 =

Answer 1

[tex]\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+.....+\frac{4}{2004\cdot 2008}=\\ \\ \frac{5-1}{1\cdot5}+\frac{9-5}{5\cdot9}+....+\frac{2008-2004}{2004\cdot2008}=\\ \frac{5}{1\cdot5}-\frac{1}{5}+\frac{9}{5\cdot9}-\frac{5}{5\cdot9}+....+\frac{2008}{2004\cdot2008}-\frac{2004}{2004\cdot2008}=\\ 1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}+....+\frac{1}{2004}-\frac{1}{2008}=\\ 1-\frac{1}{2008}=\frac{2007}{2008}[/tex]