\(B=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{20}\left(1+2+...+20\right)\)
\(\Rightarrow B=1+\dfrac{1}{2}.2.3\div2+\dfrac{1}{3}.3.4\div2+...+\dfrac{1}{20}.20.21\div2\)
\(\Rightarrow B=\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{21}{2}\)
\(\Rightarrow B=\dfrac{2+3+4+...+21}{2}\)
\(\Rightarrow B=\dfrac{230}{2}\)
\(\Rightarrow B=115\)
Vậy \(B=115\)