[tex]\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + (...) + \frac{1}{99\cdot100} =[/tex]
Znajdźmy sumy kolejnych ułamków:
[tex]\frac{1}{1\cdot2} + \frac{1}{2\cdot3} = \frac{1}{2}+\frac{1}{6} = \frac{3}{6}+\frac{1}{6} = \frac{4}{6}=\frac{2}{3}\\\\\frac{1}{1\cdot2} + \frac{1}{2\cdot3}+\frac{1}{3\cdot4} = \frac{2}{3}+\frac{1}{3\cdot4} = \frac{2}{3}+\frac{1}{12} = \frac{8}{12}+\frac{1}{12} = \frac{9}{12} = \frac{3}{4}\\\\\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \frac{1}{4\cdot5} = \frac{3}{4}+\frac{1}{4\cdot5} = \frac{3}{4}+\frac{1}{20} = \frac{15}{20}+\frac{1}{20} = \frac{16}{20} = \frac{4}{5}[/tex]
Możemy zauważyć, że:
[tex]\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + (...) + \frac{1}{n(n+1)} = \frac{n}{n+1}, \ wiec\\\\\frac{1}{1\cdot2}+\frac{1}{2\cdot3} + \frac{1}{3\cdot4} + (...) + \frac{1}{99\cdot100} = \frac{99}{100}[/tex]
