Ta có:1+2+3+...+n=\(\frac{n.\left(n+1\right)}{2}\)
=>B=\(\frac{2.2016}{\frac{2}{2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2016.2017}}\)
=>B=\(\frac{2016}{\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2016.2017}}\)
=>B=\(\frac{2016}{\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+...+\left(\frac{1}{2016}-\frac{1}{2017}\right)}\)
=>B=