a: \(A=\dfrac{1}{3\cdot9}+\dfrac{1}{9\cdot15}+...+\dfrac{1}{597\cdot603}+\dfrac{9448}{603}\)
\(=\dfrac{1}{6}\left(\dfrac{6}{3\cdot9}+\dfrac{6}{9\cdot15}+...+\dfrac{6}{597\cdot603}\right)+\dfrac{9448}{603}\)
\(=\dfrac{1}{6}\left(\dfrac{1}{3}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{15}+...+\dfrac{1}{597}-\dfrac{1}{603}\right)+\dfrac{9448}{603}\)
\(=\dfrac{1}{6}\left(\dfrac{1}{3}-\dfrac{1}{603}\right)+\dfrac{9448}{603}\)
\(=\dfrac{1}{6}\cdot\dfrac{200}{603}+\dfrac{9448}{603}=\dfrac{100+9448\cdot3}{3\cdot603}=\dfrac{28444}{1809}\)
b: \(B=\dfrac{1}{1\cdot9}+\dfrac{1}{9\cdot17}+...+\dfrac{1}{793\cdot801}\)
\(=\dfrac{1}{8}\left(\dfrac{8}{1\cdot9}+\dfrac{8}{9\cdot17}+...+\dfrac{8}{793\cdot801}\right)\)
\(=\dfrac{1}{8}\cdot\left(1-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+...+\dfrac{1}{793}-\dfrac{1}{801}\right)\)
\(=\dfrac{1}{8}\left(1-\dfrac{1}{801}\right)=\dfrac{1}{8}\cdot\dfrac{800}{801}=\dfrac{100}{801}\)
