\(\dfrac{2018}{6}+\dfrac{2018}{12}+...+\dfrac{2018}{2017.2018}\\ =2018\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2017.2018}\right)\\ =2018\left(\dfrac{1}{2}-\dfrac{1}{2018}\right)\\ =1009\)
Đặt \(A=\dfrac{2018}{6}+\dfrac{2018}{12}+\dfrac{2018}{20}+..+\dfrac{2018}{2017.2018}\)
\(A=2018\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{2017.2018}\right)\)
\(A=2018\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2017.2018}\right)\)
\(A=2018\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2017}-\dfrac{1}{2018}\right)\)
\(A=2018\left(\dfrac{1}{2}-\dfrac{1}{2018}\right)\)
\(A=2018\left(\dfrac{1006}{2018}-\dfrac{1}{2018}\right)=2018.\dfrac{1005}{2018}=1005\)
