\(\dfrac{5}{1.3}+\dfrac{5}{3.5}+\dfrac{5}{5.7}+...+\dfrac{5}{99.101}\) \(=A\)
\(A=5.\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.100}\right)\)
\(2A=5.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\right)\)
\(2A=5.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(2A=5.\left(1-\dfrac{1}{101}\right)\)
\(2A=5.\dfrac{100}{101}=\dfrac{500}{101}\)
\(A=\dfrac{500}{101}:2=\dfrac{250}{101}\)
ta có
\(\dfrac{5}{1.3}+\dfrac{5}{3.5}+\dfrac{5}{5.7}+....+\dfrac{5}{99.101}=\dfrac{5}{2}\left[\left(\dfrac{1}{1}-\dfrac{1}{3}\right)+\left(\dfrac{1}{3}-\dfrac{1}{5}\right)+....+\left(\dfrac{1}{97}-\dfrac{1}{99}\right)+\left(\dfrac{1}{99}-\dfrac{1}{101}\right)\right]\)
\(=\dfrac{5}{2}\left(1-\dfrac{1}{101}\right)=\dfrac{5}{2}.\dfrac{100}{101}=\dfrac{250}{101}\)
\(\dfrac{5}{1.3}+\dfrac{5}{3.5}+\dfrac{5}{5.7}+...+\dfrac{5}{99.101}\)
\(=\dfrac{5}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=\dfrac{5}{2}\left(1-\dfrac{1}{101}\right)\)
\(=\dfrac{250}{101}\)
