Answer     35910

This expression can be computed using hockey stick identity, as following: $$\begin{array}{rl} & 1\times 2\times 3 + 2\times 3\times 4 + \cdots + 18\times 19\times 20 \\ = &\displaystyle (1\times 2\times 3) \times \left(\frac{1\times 2\times 3}{1\times 2\times 3} + \frac{2\times 3\times 4}{1\times 2\times 3} + \cdots + \frac{18\times 19\times 20}{1\times 2\times 3}\right) \\ = & 6\times\left(\displaystyle\binom{3}{3} + \binom{4}{3} + \cdots + \binom{20}{3} \right) \\ = & 6 \times \displaystyle\binom{21}{4} \\ = &\boxed{35910} \end{array}$$

Alternatively, this problem can also be solved by transforming the original sequence to the sum of power series. Firstly, we note that $$n(n+1)(n+2) = n^3 + 3n^2 + 2n$$

Therefore $$\begin{array}{rl} & 1\times 2 \times 3 + 2\times 3\times 4 + \cdots + 18\times 19\times 20 \\ = & (1^3 + 3\times 1^2 + 2\times 1) + (2^3 + 3\times 2^2 + 2\times 2) + \cdots + (18^3 + 3\times 18^2 + 2\times 18) \\ = & (1^3 + 2^3 + \cdots + 18^3) + 3\times (1^2 + 2^2 + \cdots + 18^2) + 2\times (1 + 2 + \cdots + 18) \\ = & \Big(\frac{18\times (18 + 1)}{2}\Big)^2 + 3\times \frac{18\times (18+1) \times (2\times 18+1)}{6} + 2\times \frac{18\times (18+1)}{2} \\ = &\boxed{35910}\end{array}$$