Question

Answer 1

\(\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\\ =\dfrac{1}{3\cdot6}+\dfrac{1}{6\cdot9}+...+\dfrac{1}{30\cdot33}\\ =\dfrac{1}{3}\cdot\left(\dfrac{3}{3\cdot6}+\dfrac{3}{6\cdot9}+...+\dfrac{3}{30\cdot33}\right)\\ =\dfrac{1}{3}\cdot\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\\ =\dfrac{1}{3}\cdot\left(\dfrac{1}{3}-\dfrac{1}{33}\right)\\ =\dfrac{1}{3}\cdot\dfrac{11-1}{33}\\ =\dfrac{10}{99}\)