Question

Решите пожайлуста.....

Answer 1

$1)~~~\displaystyle \frac{1}{1\cdot2}+ \frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}=\\\\ =\frac{2-1}{1\cdot2}+ \frac{3-2}{2\cdot3}+\frac{4-3}{3\cdot4}+...+\frac{100-99}{99\cdot100}=\\ \\ =1- \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} +...+ \frac{1}{99}- \frac{1}{100} =1- \frac{1}{100} = \frac{99}{100}$


$2)~~~ \displaystyle \frac{1}{2\cdot4} +\frac{1}{4\cdot6} +\frac{1}{6\cdot8} +...+\frac{1}{98\cdot100} =\\ \\ \\ = \frac{1}{2} \cdot \bigg( \frac{1}{1\cdot2}+ \frac{1}{2\cdot3}+ \frac{1}{3\cdot4}+...+ \frac{1}{49\cdot50}\bigg)=\\ \\ \\ = \frac{1}{2}\cdot\bigg(1- \frac{1}{2} + \frac{1}{2} -\frac{1}{3} +\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50} \bigg)= \frac{1}{2} \cdot\bigg(1-\frac{1}{50}\bigg)= \frac{49}{100}$