\(\dfrac{6}{1\cdot3}+\dfrac{6}{3\cdot5}+...+\dfrac{6}{\left(n-2\right)n}\\ =3\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(n-2\right)n}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{n-2}-\dfrac{1}{n}\right)\\ =3\left(1-\dfrac{1}{n}\right)\\ =3\cdot\dfrac{n-1}{n}\)