Ta có: $\dfrac{2}{n(n+1)(n+2)}=\dfrac{n+2-n}{n(n+1)(n+2)}=\dfrac{1}{n(n+1)}-\dfrac{1}{(n+1)(n+2)}$
$\Rightarrow B= \dfrac{1}{6}+\dfrac{1}{24}+\dfrac{1}{60}+...+\dfrac{1}{990}
\\=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{9.10.11}
\\=\dfrac{1}{2}(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{9.10.11})
\\=\dfrac{1}{2}(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{9.10}-\dfrac{1}{10.11})
\\=\dfrac{1}{2}(\dfrac{1}{1.2}-\dfrac{1}{10.11})=\dfrac{1}{2}.\dfrac{27}{55}=\dfrac{27}{110}$