\(\dfrac{5}{1\cdot3}+\dfrac{5}{3\cdot5}+\dfrac{5}{5\cdot7}+...+\dfrac{5}{201\cdot203}\)
= \(\dfrac{5}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{201\cdot203}\right)\)
= \(\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{201}-\dfrac{1}{203}\right)\)
= \(\dfrac{5}{2}\left(1-\dfrac{1}{203}\right)\)
= \(\dfrac{5}{2}\cdot\dfrac{202}{203}=\dfrac{505}{203}\)
Ta có :
\(\dfrac{5}{1.3}+\dfrac{5}{3.5}+\dfrac{5}{5.7}+...+\dfrac{5}{201.203}\)
\(=\dfrac{5}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{201.203}\right)\)
\(=\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-...+\dfrac{1}{201}-\dfrac{1}{203}\right)\)
\(=\dfrac{5}{2}\left(1-\dfrac{1}{203}\right)\)
\(=\dfrac{5}{2}.\dfrac{202}{203}\)
\(=\dfrac{505}{203}\)