\(\frac{1}{5.7}+\frac{1}{7.9}+....+\frac{1}{2015.2017}=\frac{2}{2}\left(\frac{1}{5.7}+\frac{1}{7.9}+....+\frac{1}{2015.2017}\right)\)
\(=\frac{1}{2}\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{2015.2017}\right)=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{1017}\right)=\frac{1006}{10085}\)
\(\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2015.2017}\)
\(=>2\left(\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2015.2017}\right)\)\(:2\)
\(=>\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{2015.2017}\right):2\)
\(=>\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2015}-\frac{1}{2017}\right):2\)
\(=>\left(\frac{1}{5}-\frac{1}{2017}\right):2\)
\(=>-4,7:2=-2,35\)
Gọi tổng trên là S . Ta có :
\(S=\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2015.2017}\)
\(\Rightarrow2S=2.\left(\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2015.2017}\right)\)
\(\Rightarrow2S=\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{2015.2017}\)
\(\Rightarrow2S=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2015}-\frac{1}{2017}\)
\(\Rightarrow2S=\frac{1}{5}-\frac{1}{2017}\)
\(\Rightarrow2S=\frac{2012}{10085}\)
\(\Rightarrow S=\frac{2012}{10085}:2\)
\(\Rightarrow S=\frac{1006}{10085}\)
Vậy tổng trên là \(\frac{1006}{10085}\)