Question

A=2^2/1*3+2^2/3*5+...+2^2/99*101

Answer 1

A= \(\dfrac{2^2}{1.3}+\dfrac{2^2}{3.5}+...\dfrac{2^2}{99.101}\)

= \(\dfrac{4}{1.3}+\dfrac{4}{3.5}+...+\dfrac{4}{99.101}\)

= \(\dfrac{4}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

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

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

= \(2.\dfrac{100}{101}\)

= \(\dfrac{200}{101}\)