A or B 2???

$\begin{aligned} B & = 2 + 4 + 6 + \cdots + 2020 \\ A & = 1 + 3 + 5 + \cdots + 2019 + 2021 \\ \implies B - A & = \underbrace{1 + 1 + 1 + \cdots + 1}_{1010 \times 1} - 2021 = 1010 - 2021 = - 1011 \end{aligned}$

Therefore, $\boxed{A>B}$ .

I found a pattern for both A and B.

I applied them, and for A, I got $1011^{1011}$ = 1 022 121.

For B, I got $1010 \times 1011$ = 1 021 110.

Obviously, 1 022121 > 1 021 110 .

Therefore, $\boxed {A > B}$

Each number in $B$ is $1$ higher than some number in $A$ and these pair off with each number in $A$ except $2021$ , so $B > A - 2021$ .

Next, note that there are $1010$ of these pairs, each making $B$ larger by $1$ , so we get $B = A - 2021 + 1010$ .

This boils down to $B = A - 1011$ or just $B < A$ .

We can write,

$A=1+3+5+...+2021=1010²+2021$

$B=2+4+6+...+2020=1010²+1010$

So, $\boxed{A>B}$