Question

F=1/18+1/54+1/108+...+1/990

Answer 1

\(F=\dfrac{1}{3\cdot6}+\dfrac{1}{6\cdot9}+\dfrac{1}{9\cdot12}+...+\dfrac{1}{30\cdot33}\)

\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{10}{33}=\dfrac{10}{99}\)

Answer 2

\(3F=\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{10}-\dfrac{1}{33}\) \(\Rightarrow\) \(3F=\dfrac{1}{3}-\dfrac{1}{33}=\dfrac{30}{99}\) \(\Rightarrow\) \(F=\dfrac{10}{99}\)