Question

Cho biểu thức A bằng \(\dfrac{-10}{56}+\dfrac{-10}{140}+\dfrac{-10}{260}+...+\dfrac{-10}{1400}sosánhAvới\dfrac{1}{3}\)

Answer 1

\(A=-10\left(\dfrac{1}{56}+\dfrac{1}{140}+...+\dfrac{1}{1400}\right)\)

\(\Leftrightarrow2\cdot A=-10\cdot\left(\dfrac{1}{28}+\dfrac{1}{70}+...+\dfrac{1}{700}\right)\)

=>\(A=-5\left(\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+...+\dfrac{1}{25\cdot28}\right)\)

\(=\dfrac{-5}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)

=-5/3*3/14=-5/14<1/3