Question

ПОМОГИТЕ ПОЖАЛУЙСТА. ТЕМА ДРОБИ​

Answer 1

$\frac{1}{2\cdot4} +\frac{1}{4\cdot6} +...+\frac{1}{2n(2n+2)} = \frac{1}{2} \cdot (\frac{2}{2\cdot4} +\frac{2}{4\cdot6} +...+\frac{2}{2n(2n+2)}) = \frac{1}{2} \cdot (\frac{4-2}{2\cdot4} +\frac{6-4}{4\cdot6} +...+\frac{2n+2-2n}{2n(2n+2)}) = \frac{1}{2} \cdot (\frac{4}{2\cdot4} -\frac{2}{2\cdot4}+\frac{6}{4\cdot6}-\frac{4}{4\cdot6} +...+\frac{2n+2}{2n(2n+2)}-\frac{2n}{2n(2n+2)}) =$

$\frac{1}{2} \cdot (\frac{1}{2} -\frac{1}{4}+\frac{1}{4}-\frac{1}{6} +...+\frac{1}{2n}-\frac{1}{2n+2})=\frac{1}{2} \cdot (\frac{1}{2} -\frac{1}{2n+2}) = \frac{1}{2} \cdot (\frac{1}{2} -\frac{1}{2(n+1)}) = \frac{1}{2} \cdot (\frac{n+1}{2(n+1)} -\frac{1}{2(n+1)})= \frac{1}{2} \cdot \frac{n+1-1}{2(n+1)} = \frac{n}{4(n+1)} \\E)$