Đặt A= \(\frac{4}{1.2}+\frac{4}{2.3}+...+\frac{4}{2014.2015}\)
=>\(\frac{1}{4}A=\frac{1}{4}\left(\frac{4}{1.2}+\frac{4}{2.3}+...+\frac{4}{2014.2015}\right)\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2014.2015}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2014}-\frac{1}{2015}\)
=\(1-\frac{1}{2015}\)
=\(\frac{2014}{2015}\)
vậy ....
dấu "." là dấu nhân
\(\frac{4}{1\cdot2}+\frac{4}{2\cdot3}+\frac{4}{3\cdot4}+...+\frac{4}{2014\cdot2015}\)
\(=4\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2014\cdot2015}\right)\)
\(=4\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\right)\)
\(=4\left(1-\frac{1}{2015}\right)\)
\(=4\cdot\frac{2014}{2015}\)
\(=\frac{8056}{2015}\)
