\(S=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2017.2019}\)
\(\Leftrightarrow S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2017.2019}\)
\(\Rightarrow2S=2\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2017.2019}\right)\)
\(\Leftrightarrow2S=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2017.2019}\)
\(\Leftrightarrow2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{2}{2019}\)
\(\Leftrightarrow2S=1-\frac{1}{2019}=\frac{2018}{2019}\)
\(\Rightarrow S=\frac{2018}{2019}:2=\frac{1009}{2019}\)
Vậy \(S=\frac{1009}{2019}.\)
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2017.2019}\)
\(\Rightarrow S=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2017.2019}\)
\(\Rightarrow S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2017}-\frac{1}{2019}\)
\(\Rightarrow S=\frac{2018}{2019}\)
Ta có : \(S=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+.....+\frac{1}{2017.2019}\)
\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2017.2019}\)
=> \(2S=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{2017.2019}\)
=> \(2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{2017}-\frac{1}{2019}\)
=> \(2S=1-\frac{1}{2019}=\frac{2018}{2019}\)
=> \(S=\frac{2018}{2019}.\frac{1}{2}=\frac{1009}{2019}\)
