Ta có: \(2S=\dfrac{2}{1\cdot3}+\dfrac{2}{2\cdot4}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{7\cdot9}+\dfrac{2}{8\cdot10}\)
\(\Leftrightarrow2S=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{8}-\dfrac{1}{10}\)
\(\Leftrightarrow2S=1+\dfrac{1}{2}-\dfrac{1}{9}-\dfrac{1}{10}=\dfrac{58}{45}\)
\(\Rightarrow S=\dfrac{29}{45}\)
Ta có:
\(S=\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+...+\dfrac{1}{7.9}+\dfrac{1}{8.10}\)
\(=\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{7.9}\right)\) \(+\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{8.10}\right)\)
Đặt \(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{7.9}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{7.9}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{7}-\dfrac{1}{9}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{9}\right)=\dfrac{4}{9}\)
Đặt \(B=\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{8.10}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{8.10}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{8}-\dfrac{1}{10}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=\dfrac{1}{5}\)
\(\Rightarrow S=A+B=\dfrac{4}{9}+\dfrac{1}{5}=\dfrac{29}{45}\)
Vậy \(S=\dfrac{29}{45}\)
