\(B=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\)
\(B=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(B=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(B=1-\dfrac{1}{100}\)
\(B=\dfrac{99}{100}\)
Vậy \(B=\dfrac{99}{100}\)
B \(=\) \(\dfrac{1}{2}\) \(+\) \(\dfrac{1}{6}\) \(+\) \(\dfrac{1}{12}\) \(+\) \(\dfrac{1}{20}\) \(+\) \(\dfrac{1}{30}\) \(+\) . . . . . \(+\) \(\dfrac{1}{9900}\)
\(=\) \(\dfrac{1}{1.2}\) \(+\) \(\dfrac{1}{2.3}\) \(+\) \(\dfrac{1}{3.4}\) \(+\) \(\dfrac{1}{4.5}\) \(+\) \(\dfrac{1}{5.6}\) \(+\) . . . . . \(+\) \(\dfrac{1}{99.100}\)
\(=\) \(\dfrac{1}{1}\) \(-\) \(\dfrac{1}{2}\) \(+\) \(\dfrac{1}{2}\) \(-\) \(\dfrac{1}{3}\) \(+\) \(\dfrac{1}{3}\) \(-\) \(\dfrac{1}{4}\) \(+\) \(\dfrac{1}{4}\) \(-\) \(\dfrac{1}{5}\) \(+\) \(\dfrac{1}{5}\) \(-\) \(\dfrac{1}{6}\) \(+\) . . . . . \(+\) \(\dfrac{1}{99}\) \(-\) \(\dfrac{1}{100}\)
\(=\) \(\dfrac{1}{1}\) \(-\) \(\dfrac{1}{100}\)
\(=\) \(\dfrac{99}{100}\)
