Ta có 1+2+....+n = n(n+1):2
=>\(\frac{1}{1+2+....+n}=\frac{2}{n\left(n+1\right)}\)
\(A=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+.......+\frac{2}{2014.2015}\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{2014.2015}\right)\)
\(=2\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2014}-\frac{1}{2015}\right)\)
\(=2.\left(1-\frac{1}{2015}\right)=\frac{2.2014}{2015}=\frac{4028}{2015}\)