Question

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

Answer 1

A=\(\dfrac{3}{3.5}+\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{99.101}\)

A=\(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

A=\(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{101}\right)\)

A=\(\dfrac{3}{2}.\dfrac{98}{303}\)

A=\(\dfrac{49}{101}\)