This problem can be solved by using the hockey stick identity. $$\begin{array}{rl} &1\times 2 + 2\times 3 + 3\times 4 + \cdots + 2015 \times 2016\\ =\quad& (2!)\cdot\Big(\frac{1\times 2}{2!} + \frac{2\times 3}{2!} + \frac{3\times 4}{2!} + \cdots + \frac{2015 \times 2016}{2!}\Big)\\ =\quad&2\times(C_2^2+C_3^2+C_4^2+\cdots+C_{2016}^2)\\ =\quad&2\times C_{2017}^3\\ =\quad& 2731179306 \end{array}$$