\(M=\dfrac{3^2}{2.5}+\dfrac{3^2}{5.8}+...+\dfrac{3^2}{98.101}\\ =3\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{98.101}\right)\\ =3\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{98}-\dfrac{1}{101}\right)\\ =3.\left(\dfrac{1}{2}-\dfrac{1}{101}\right)\\ =3.\dfrac{99}{202}\\ =\dfrac{297}{202}\)