Question

Cho A= \(\dfrac{4}{2.4}\)+\(\dfrac{4}{4.6}\)+\(\dfrac{4}{6.8}\)+...+\(\dfrac{4}{46.48}\)+\(\dfrac{4}{48.50}\)

Answer 1

\(A=2\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{48.50}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{50}\right)\)

\(=2\times\dfrac{12}{25}=\dfrac{24}{25}\)

Answer 2

\(=>A=4.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{46}-\dfrac{1}{48}+\dfrac{1}{48}-\dfrac{1}{50}\right)\)

\(A=4.\left(\dfrac{1}{2}-\dfrac{1}{50}\right)=4.\left(\dfrac{25}{50}-\dfrac{1}{50}\right)=\dfrac{4.24}{50}=\dfrac{48}{25}\)