\(B=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+..+\dfrac{1}{98.99.100}\)
\(\dfrac{1}{1.2}-\dfrac{2}{2.3}=\dfrac{2}{1.2.3};...;\dfrac{1}{98.99}-\dfrac{1}{99.100}=\dfrac{2}{98.99.100}\)
=> 2B = \(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+.....+\dfrac{2}{98.99.100}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+..+\dfrac{1}{98.99}-\dfrac{1}{99.100}\)
=> 2B = \(\dfrac{1}{2}-\dfrac{1}{99.100}=\dfrac{4949}{9900}\)
=> B = \(\dfrac{4949}{19800}\)