\(\dfrac{6}{1.3}+\dfrac{6}{3.5}+...+\dfrac{6}{99.100}\\ =3\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.100}\right)\\ =3\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =3\left(1-\dfrac{1}{100}\right)\\ =3.\dfrac{99}{100}\\ =\dfrac{297}{100}\)