Odd-Even Number Sum Series

Rewrite the equation

$(2-1) + (4-3) + ... + (2022-2021) \\ = \sum\limits_{n=1}^{1011} 2n-(2n-1) \\ =\sum\limits_{n=1}^{1011} 1 \\ = 1011$

We can rewrite the equation as:

$(-1010\times 2-1)+(-1009\times 2 -1)+(-1008\times 2-1)+...+(0\times 2 - 1)+2\times 1+2\times 2+...+2\times 1011$

Simplifying, we get: $\displaystyle -1\times 1011+(-2\sum^{1010}_{n\ =\ 1}n)+(2\sum^{1011}_{n\ =\ 1}{n})$

(Explanation of 1st term: there are 1011 terms from $0 \times 2$ to $-1010 \times 2$ [as $1010-0+1=1011$ ], and the term $-1$ appears in all 1011 terms)

Following that, $-1\times 1011+2022$

And $-1011+2022$

Finally, we get $\boxed{1011}$