b)Đặt A=\(\dfrac{1}{2.4}\)+\(\dfrac{1}{4.6}\)+...+\(\dfrac{1}{2016.2018}\)
2A=\(\dfrac{2}{2.4}\)+\(\dfrac{2}{4.6}\)+...+\(\dfrac{2}{2016.2018}\)
2A=\(\dfrac{1}{2}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{6}\)+...+\(\dfrac{1}{2016}\)-\(\dfrac{1}{2018}\)
2A=\(\dfrac{1}{2}\)-\(\dfrac{1}{2018}\)
2A=\(\dfrac{504}{1009}\)
⇒A=\(\dfrac{252}{1009}\)