Đặt A = \(\dfrac{2}{5}+\dfrac{2}{10}+\dfrac{2}{20}+...+\dfrac{2}{320}\)
= 2 x ( \(\dfrac{1}{5}+\dfrac{1}{10}+\dfrac{1}{20}+...+\dfrac{1}{320}\) )
Đặt B = \(\dfrac{1}{5}+\dfrac{1}{10}+\dfrac{1}{20}+...+\dfrac{1}{320}\)
\(\dfrac{1}{2}\) x B = \(\dfrac{1}{2}\) x ( \(\dfrac{1}{5}+\dfrac{1}{10}+\dfrac{1}{20}+...+\dfrac{1}{320}\) )
= \(\dfrac{1}{10}+\dfrac{1}{20}+\dfrac{1}{40}+...+\dfrac{1}{640}\)
B - \(\dfrac{1}{2}\) x B = ( \(\dfrac{1}{5}+\dfrac{1}{10}+\dfrac{1}{20}+...+\dfrac{1}{320}\) ) - ( \(\dfrac{1}{10}+\dfrac{1}{20}+\dfrac{1}{40}+...+\dfrac{1}{640}\))
B x ( 1 - \(\dfrac{1}{2}\) ) = \(\dfrac{1}{5}-\dfrac{1}{640}\)
\(\dfrac{1}{2}\) x B = \(\dfrac{127}{640}\)
B = \(\dfrac{127}{640}:\dfrac{1}{2}=\dfrac{127}{320}\)
Vậy A = 2 x \(\dfrac{127}{320}=\dfrac{127}{160}\)
