\(A=\frac{3}{1\cdot3}+\frac{3}{2\cdot5}+\frac{3}{5\cdot7}+...+\frac{3}{2015\cdot2017}\)
\(\frac{2}{3}A=\frac{2}{3}\left(\frac{3}{1\cdot3}+\frac{3}{3\cdot5}+\frac{3}{5\cdot7}+...+\frac{3}{2015\cdot2017}\right)\)
\(\frac{2}{3}A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{2015\cdot2017}\)
\(\frac{2}{3}A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\)
\(\frac{2}{3}A=1-\frac{1}{2017}\)
\(\frac{2}{3}A=\frac{2016}{2017}\)
\(A=\frac{2016}{2017}:\frac{2}{3}\)
\(A=\frac{2016}{2017}\cdot\frac{3}{2}\)
\(A=1,499256321\)
A x 2/3=2/3 x(3/1x3+3/3x5+...+3/2015x2017)
A x 2/3=2/1x3+2/3x5+...+2/2015x2017
A x 2/3=1-1/3+1/3-1/5+...+1/2015-1/2017
A x 2/3=1-1/2017
A x 2/3=2016/2017
A = 3024/2017
đề đúng
\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{2015.2017}+\frac{3}{2017.2019}\)
\(A=\frac{3}{2}\cdot\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2017.2019}\right)\)
\(A=\frac{3}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)
\(A=\frac{3}{2}\cdot\left(1-\frac{1}{2019}\right)\)
\(A=\frac{3}{2}\cdot\frac{2018}{2019}\)
\(A=\frac{1009}{673}\)
