Question

Cho 2 biểu thức D và E như sau

\(D=\dfrac{10}{56}+\dfrac{10}{140}+\dfrac{10}{260}+...+\dfrac{10}{1400}\)

\(E=\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+3+4+...+24}\)

Tính \(\dfrac{D}{E}\)

Answer 1

\(D=\dfrac{5}{4.7}+\dfrac{5}{7.10}+\dfrac{5}{10.13}+...+\dfrac{5}{25.28}\)

\(=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{28}\right)=\dfrac{5}{3}.\dfrac{6}{28}=\dfrac{5}{14}\)

\(E=\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{24.25}=2\left(\dfrac{1}{2}-\dfrac{1}{25}\right)=\dfrac{2.23}{50}=\dfrac{23}{25}.\)

\(\dfrac{D}{E}=\dfrac{5}{24}.\dfrac{25}{23}=\dfrac{125}{552}.\)