Question

Помогите пожалуйстааа

Answer 1

$\boxed {\; \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\; }$

$\frac{2}{1\cdot 2}+\frac{2}{2\cdot 3}+\frac{2}{3\cdot 4}+\frac{2}{4\cdot 5}+\frac{2}{5\cdot 6}+\frac{2}{6\cdot 7}+\frac{2}{7\cdot 8}=\\\\=2\cdot \Big (\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\frac{1}{4\cdot 5}+\frac{1}{5\cdot 6}+\frac{1}{6\cdot 7}+\frac{1}{7\cdot 8}\Big )=\\\\=2\cdot \Big ((\frac{1}{1}-\frac{1}2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{6})+(\frac{1}{6}-\frac{1}{7})+(\frac{1}{7}-\frac{1}{8})\Big )=\\\\=2\cdot \Big(1-\frac{1}{8}\Big )=2\cdot \frac{7}{8}=\frac{7}{4}=1\frac{3}{4}$