Ta có:\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.............+\frac{2}{99.101}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+............+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
Đặt \(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(\Rightarrow A=2\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\right)\)
Đặt \(B=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\)
\(\Rightarrow2B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow2B=1-\frac{1}{100}=\frac{99}{100}\)
\(\Rightarrow B=\frac{99}{100}:2\)
Ta có : \(A=2B=\frac{99}{100}:2.2=\frac{99}{100}\)
Vậy tổng sau bằng \(\frac{99}{100}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(\Leftrightarrow\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...\frac{1}{99.101}\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(\Leftrightarrow1-\frac{1}{101}=\frac{100}{101}\)