Ta có: \(A=\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+...+\frac{4}{2014.2015}\)
\(=1\left(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{2014.2015}\right)\)
\(=1\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\right)\)
\(=1\left(1-\frac{1}{2015}\right)\)
\(=1\left(\frac{2015}{2015}-\frac{1}{2015}\right)-1\left(\frac{2014}{2015}\right)=\frac{2014}{2015}\)
Vậy.....
\(A=\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+....+\frac{4}{2014.2015}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2014}-\frac{1}{2015}\)
\(A=\frac{1}{1}-\frac{1}{2015}=\frac{2015}{2015}-\frac{1}{2015}=\frac{2014}{2015}\)
Ta có:
A = \(\frac{4}{1.2}+\frac{4}{2.3}+...+\frac{4}{2014.2015}\)
=\(4.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2014.2015}\right)\)
=\(4.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2014}-\frac{1}{2015}\right)\)
=\(4.\left(\frac{1}{1}-\frac{1}{2015}\right)\)
=\(4.\frac{2014}{2015}\)
=\(\frac{8056}{2015}\)
BẠN CỘNG TÁC VIÊN TTH LÀM SAI RỒI NHÉ
\(\frac{4}{1.2}+\frac{4}{2.3}+...+\frac{4}{2015.2015}\)
\(=4\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2014.2015}\right)\)
\(=4\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{2014}+\frac{1}{2014}-\frac{1}{2015}\right)\)
\(=4\left(1-\frac{1}{2015}\right)=\frac{4.2014}{2015}=\frac{8056}{2015}\)