A = \(\left(1-\dfrac{1}{1+2}\right)\left(1-\dfrac{1}{1+2+3}\right)...\left(1-\dfrac{1}{1+2+3+...+2017}\right)\)
= \(\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)...\left(1-\dfrac{2}{2017.2018}\right)\)
= \(\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}...\dfrac{2015.2018}{2016.2017}.\dfrac{2016.2019}{2017.2018}\)
= \(\dfrac{2019}{3.2017}=\dfrac{673}{2017}\)