\(\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{2499}\\ =\dfrac{1}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}+\dfrac{1}{51}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{51}\right)=\dfrac{1}{2}.\dfrac{16}{51}\\ =\dfrac{8}{51}\)