\(\dfrac{5}{3}+\dfrac{5}{6}+\dfrac{5}{10}+\dfrac{5}{15}+\dfrac{5}{21}+...+\dfrac{5}{4950}=\dfrac{10}{6}+\dfrac{10}{12}+\dfrac{10}{20}+\dfrac{10}{30}+\dfrac{10}{42}+...+\dfrac{10}{9900}=10.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{99.100}\right)=10.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=10.\left(\dfrac{1}{2}-\dfrac{1}{100}\right)=10.\dfrac{49}{100}=\dfrac{49}{10}\)