\(B=\dfrac{4}{2.4}+\dfrac{4}{4.6}+...+\dfrac{4}{98.100}\)
\(\Rightarrow5B=\dfrac{20}{2.4}+\dfrac{20}{4.6}+...+\dfrac{20}{98.100}\)
\(\Rightarrow5B=10\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{98.100}\right)\)
\(\Rightarrow5B=10\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\)
\(\Rightarrow5B=10\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\)
\(\Rightarrow5B=10.\dfrac{49}{100}\)
\(\Rightarrow5B=\dfrac{49}{10}\)
Vậy \(5B=\dfrac{49}{10}\)
Ta có: B = \(\dfrac{4}{2.4}\) + \(\dfrac{4}{4.6}\) + \(\dfrac{4}{6.8}\) + ... + \(\dfrac{4}{98.100}\).
=> \(\dfrac{B}{2}\) = \(\dfrac{2}{2.4}\) + \(\dfrac{2}{4.6}\) + \(\dfrac{2}{6.8}\) + ... + \(\dfrac{2}{98.100}\)
=\(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{8}\) + ... + \(\dfrac{1}{98}\) - \(\dfrac{1}{100}\)
= \(\dfrac{1}{2}\) - \(\dfrac{1}{100}\) = \(\dfrac{49}{100}\).
=> B = \(\dfrac{49}{200}\).
=> 5B = \(\dfrac{49}{200}\) . 5 = \(\dfrac{49}{40}\).
Vậy 5B = \(\dfrac{49}{40}\).
