Question

 

A=\(\frac{2x2019}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+....+\frac{1}{1+1+3+4+.....+2019}}\)

Các bạn tìm A giúp mk nha! mk  bí quá rồi mai phải có bài.giups mk nha.thanks nhiều(ai đúng mk tick nha)

Answer 1

Đặt: \(B=1+\frac{1}{1+2}+\frac{1}{1+2+3}+........+\frac{1}{1+2+3+........+2019}\)

Ta có: \(1+2=\frac{2.3}{2}\); \(1+2+3=\frac{3.4}{2}\); .............. ; \(1+2+3+......+2019=\frac{2019.2020}{2}\)

\(\Rightarrow B=\frac{2}{2}+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+........+\frac{1}{\frac{2019.2020}{2}}\)

\(=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+......+\frac{2}{2019.2020}\)

\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{2019.2020}\right)\)

\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{2019}-\frac{1}{2020}\right)\)

\(=2.\left(1-\frac{1}{2020}\right)=2.\frac{2019}{2020}=\frac{2019}{1010}\)

\(\Rightarrow A=\frac{2.2019}{\frac{2019}{1010}}=2.1010=2020\)