Question

A=(3/3.5) + (3/5.7) + (3/7.9) +...+ (3/99.101)

Answer 1

\(A=\dfrac{3}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{99\cdot101}\right)\)

\(=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=\dfrac{3}{2}\cdot\dfrac{98}{303}=\dfrac{49}{101}\)