Đặt A = \(\dfrac{2}{1.4}+\dfrac{2}{4.7}+\dfrac{2}{7.10}+...+\dfrac{2}{91.94}+\dfrac{2}{94.97}\)
\(\dfrac{3A}{2}\)= \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{91.94}+\dfrac{3}{94.97}\)
\(\dfrac{3A}{2}=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{91}-\dfrac{1}{94}+\dfrac{1}{94}-\dfrac{1}{97}\)
\(\dfrac{3A}{2}\) = \(\dfrac{1}{1}-\dfrac{1}{97}\)
\(\dfrac{3A}{2}\) = \(\dfrac{96}{97}\)
3A = \(\dfrac{96}{97}.2\)
3A = \(\dfrac{192}{97}\)
A = \(\dfrac{192}{97}:3\)
A = \(\dfrac{64}{97}\)
Vậy A = \(\dfrac{64}{97}\)
\(\dfrac{2}{1.4}\)+\(\dfrac{2}{4.7}\)+\(\dfrac{2}{7.10}\)+....+\(\dfrac{2}{91.94}\)
=2.(1/1.4+1/4.7+1/7.10+...+1/91.94)
=2.\(\dfrac{3}{3}\).(1/1.4+1/4.7+1/7.10+...+1/91.94)
=\(\dfrac{2}{3}\).(3/1.4+3/4.7+3/7.10+...+3/91.94)
=\(\dfrac{2}{3}\).(1/1-1/4+1/4-1/7+1/7-1/10+.....1/91-1/94
=\(\dfrac{2}{3}\).(1-1/94)
=\(\dfrac{2}{3}\).93/94
=31/47
\(\dfrac{2}{1\cdot4}+\dfrac{2}{4\cdot7}+\dfrac{2}{7\cdot10}+...+\dfrac{2}{91\cdot94}+\dfrac{2}{94\cdot97}\)
\(=\dfrac{2}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{91\cdot94}+\dfrac{3}{94\cdot97}\right)\)
\(=\dfrac{2}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{91}-\dfrac{1}{94}+\dfrac{1}{94}-\dfrac{1}{97}\right)\)
\(=\dfrac{2}{3}\left(1-\dfrac{1}{97}\right)\)
\(=\dfrac{2}{3}\left(\dfrac{97}{97}-\dfrac{1}{97}\right)\)
\(=\dfrac{2}{3}\cdot\dfrac{96}{97}\)
\(=\dfrac{64}{97}\)
