Từ từ mình trình bày cho nha
\(\frac{100}{51}\)nha
HT
\(1+\frac{1}{\frac{\left(1+2\right)2}{2}}+\frac{1}{\frac{\left(1+3\right)3}{2}}+\frac{1}{\frac{\left(1+4\right)4}{2}}+...+\frac{1}{\frac{\left(1+2019\right)2019}{2}}\)
\(=1+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+...+\frac{1}{\frac{2019.2020}{2}}\)
\(=1+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2019.2020}\)
\(=1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\right)\)
\(=1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\right)\)
\(=1+2\left(\frac{1}{2}-\frac{1}{2020}\right)\)
\(=1+2.\frac{2018}{4020}\)
\(=1+\frac{2018}{2010}\)
\(=\frac{2013}{1005}\)
Mình sửa lai đoạn này cái nha:
\(1+2\left(\frac{1}{2}-\frac{1}{2020}\right)\)
\(=1+2.\frac{2018}{4040}\)
\(=1+\frac{2018}{2020}\)
\(=1+\frac{1009}{1010}\)
\(=\frac{2019}{1010}\)
