Question

Answer 1

a: \(A=\dfrac{2+\dfrac{2}{3}+\dfrac{2}{5}+...+\dfrac{2}{99}}{4+\dfrac{4}{3}+...+\dfrac{4}{99}}\)

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

\(B=\dfrac{5+\dfrac{5}{41}+\dfrac{5}{401}+\dfrac{5}{2024}}{9+\dfrac{9}{41}+\dfrac{9}{401}+\dfrac{9}{2024}}=\dfrac{5\left(1+\dfrac{1}{41}+\dfrac{1}{401}+\dfrac{1}{2024}\right)}{9\left(1+\dfrac{1}{41}+\dfrac{1}{401}+\dfrac{1}{2024}\right)}=\dfrac{5}{9}\)

\(A=\dfrac{1}{2}=\dfrac{5}{10}< \dfrac{5}{9}=B\)

=>A<B