A problem to start my 2018

Rewrite the sum as follows:

$\begin{aligned} S & = 1 \times 2 - 2 \times 3 + 3 \times 4 - 4 \times 5 + 5 \times 6 - \cdots - 2016 \times 2017 + 2017 \times 2018 \\ & = 1 \times 2 + 3(4-2) + 5(6-4) + \cdots + 2017(2018-2016) \\ & = 1 \times 2 + 3 \times 2 + 5 \times 2 + \cdots + 2017 \times 2 \\ & = 2 (\underbrace{\overbrace{1}^{\color{#3D99F6}a=}+3+5+\cdots+\overbrace{2017}^{\color{#3D99F6}l=}}_{\color{#3D99F6}n=1009 \text{ terms}}) & \small \color{#3D99F6} \text{Sum of AP: } S = \frac {n(a+l)}2 \\ & = 2 \times \frac {1009(1+2017)}2 \\ & = \boxed{2036162} \end{aligned}$

Let $S = 2018 \times 2017 - 2017 \times 2016 + 2016 \times 2015 - 2015 \times 2014 + . . . + 4 \times 3 - 3 \times 2 + 2 \times 1$

$S = 2018 \times 2017 - 2017 \times 2016 + 2016 \times 2015 - 2015 \times 2014 + . . . + 4 \times 3 - 3 \times 2 + 2 \times 1$ $S = 2017 \times (2018 - 2016) + 2015 \times (2016 - 2014) + . . . + 3 \times (4 - 2) + 1 \times (2 - 0)$ $S = 2017 \times 2 + 2015 \times 2 + . . . + 3 \times 2 + 1 \times 2$ $S = 2 \times (2017 + 2015 + 2013 + ... + 5 + 3 + 1)$ $S = 2 \times ((2017 + 1) + (2015 + 3) + ... + (1011 + 1007) + (1009))$ $S = 2 \times (504 \times 2018 + 1009)$ $S = 2 \times 1018081$ $S = 2036162$