Question

Answer 1

a: \(A=\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{380}\)

\(=2\left(\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{380}\right)\)

\(=2\left(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)\)

\(=2\cdot\dfrac{4-1}{20}=\dfrac{3}{10}\)

b: \(A=\dfrac{4}{28}+\dfrac{4}{70}+...+\dfrac{4}{700}\)

\(=\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{25}-\dfrac{1}{28}=\dfrac{1}{4}-\dfrac{1}{28}=\dfrac{6}{28}=\dfrac{3}{14}\)