Question

Va rog frumos , am nevoie de acest calcul cat mai repede ​

Answer 1

[tex]\displaystyle b=\frac1{1\cdot2}+\frac1{2\cdot3}+\frac1{3\cdot4}+\cdots+\frac1{2019\cdot2020} \\\\ =\frac{2-1}{1\cdot2}+\frac{3-2}{2\cdot3}+\frac{4-3}{3\cdot4}+\cdots+\frac{2020-2019}{2019\cdot2020}\\\\ =\frac{2}{1\cdot2}-\frac{1}{1\cdot2}+\frac{3}{2\cdot3}-\frac{2}{2\cdot3}+\frac{4}{3\cdot4}-\frac{3}{3\cdot4}+\cdots+\frac{2020}{2019\cdot2020}-\frac{2019}{2019\cdot2020}\\\\ [/tex]

[tex]\displaystyle= \frac11-\frac12+\frac12-\frac13+\frac13-\frac14+\cdots+\frac{1}{2019}-\frac{1}{2020}\\\\ =1-\frac1{2020}=\frac{2020-1}{2020}=\frac{2019}{2020}[/tex]