Ta có : \(1+2=\frac{2.3}{2}\) , \(1+2+3=\frac{3.4}{2}\) ,
\(1+2+3+4=\frac{4.5}{2}\) , ......... , \(1+2+3+4+....+2014=\frac{2014.2015}{2}\)
Suy ra : \(A=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2014.2015}\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2014.2015}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2014}-\frac{1}{2015}\right)\)
\(2\left(\frac{1}{2}-\frac{1}{2015}\right)\)
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2014}\)
\(A=\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+...+\frac{1}{\left(1+2014\right).2014:2}\)
\(A=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2014.2015}\)
\(A=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2014.2015}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{2014}-\frac{1}{2015}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{2015}\right)\)
\(A=2.\frac{1}{2}-2.\frac{1}{2015}\)
\(A=1-\frac{2}{2015}\)
\(A=\frac{2013}{2015}\)
