Gọi tổng là A
⇒ A = \(\dfrac{1}{28}+\dfrac{1}{70}+\dfrac{1}{130}+\dfrac{1}{208}+...+\dfrac{1}{3190}\)
⇒ 3A = \(3\left(\dfrac{1}{28}+\dfrac{1}{70}+\dfrac{1}{130}+\dfrac{1}{208}+...+\dfrac{1}{3190}\right)\)
⇒ 3A = \(\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+\dfrac{3}{13.16}+...+\dfrac{3}{55.58}\)
⇒ 3A = \(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}+...+\dfrac{1}{55}-\dfrac{1}{58}\)
⇒ 3A = \(\dfrac{1}{4}-\dfrac{1}{58}\) = \(\dfrac{29}{116}-\dfrac{2}{116}\) = \(\dfrac{27}{116}\)
⇒ A = \(\dfrac{27}{116}\): 3 = \(\dfrac{27}{116}\).\(\dfrac{1}{3}\) = \(\dfrac{9}{116}\)
