Question

giúp e ạ🤟🏻😎

Answer 1

a: \(A=\dfrac{2}{3}\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{2003\cdot2006}\right)\)

\(=\dfrac{2}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{2003}-\dfrac{1}{2006}\right)\)

\(=\dfrac{2}{3}\cdot\dfrac{501}{1003}=\dfrac{1002}{3009}=\dfrac{334}{1003}\)

b: \(B=5\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=5\cdot\dfrac{49}{100}=\dfrac{49}{20}\)