Solution:
For each integer $0\leq k\leq 1009$ define a $k$-pal to be any sequence of 2020 $A$s and $B$s, where the first $k$ terms are $B$, the last $k$ terms are $B$, and the middle $(2020 - 2k)$ terms form an antipalindromic sequence.

Now for any $k$, define $f(k)$ to be sum of the costs of all $k$-pals. Note that any $k$-pal can be created from a $(k + 1)$-pal by either
- (A) replacing the $B$ in position $(k + 1)$ with an $A$, or
- (B) replacing the $B$ in position $(2021 - k)$ with an $A$.

Therefore the sum of the costs of all $k$-pals formed using operation (A) is $(k + 1)\times f(k + 1)$. Similarly the sum of the costs of all $k$-pals formed using operation (B) is $(2021 - k)\times f(k + 1)$. Hence
$$f(k) = (k + 1)f(k + 1) + (2020 - k)f(k + 1) = ((k + 1) + (2020 - k))f(k + 1) = 2021f(k + 1).$$
Now we note that there are two different 1009-pals, with costs equal to 1010 and 1011 respectively. So
$$f(1009) = 1010 + 1011 = 2021.$$
Now if we use the formula $f(k) = 2021f(k + 1)$ iteratively, we get $f(1010 - i) = 2021^{i}$ for each $i = 1,2,3,\ldots$. Therefore
$$f(0) = 2021^{1010}$$
which is our final answer.

Solution:
Let $n$ be a positive integer. We will find an expression (in terms of $n$) for the sum of the costs of all antipalindromes of length $2n$. Note that a string of $A$s and $B$s of length $2n$ is an antipalindrome if and only if for each $i$, exactly one of the $i^{\mathrm{th}}$ and $(2n + 1 - i)^{\mathrm{th}}$ letters is an $A$ (and the other is a $B$).

Let $x_{1},x_{2},\ldots ,x_{2n}$ be variables. For any $1\leq a(1)< a(2)< \dots < a(k)\leq 2n$, consider the string of $A$s and $B$s of length $2n$, such that the $a(j)^{\mathrm{th}}$ letter is $A$ for all $j$ (and all the other letters are $B$). Let this string correspond to the term $t = x_{a(1)}x_{a(2)}x_{a(3)}\dots x_{a(k)}$. If $x_{i} = i$ for all $i$ then the value of $t$ is equal to the cost of its corresponding string. Now consider the expression
$$y = (x_{1} + x_{2n})(x_{2} + x_{2n - 1})\cdot \cdot \cdot (x_{n} + x_{n + 1}) = \prod_{j = 1}^{n}(x_{j} + x_{2n + 1 - j}).$$
If we expand the brackets then we get $2^{n}$ terms, each in the form $t = x_{a(1)}x_{a(2)}x_{a(3)}\dots x_{a(n)}$ such that for each $j = 1,2,\ldots ,n$ either $a(j) = j$ or $a(j) = 2n + 1 - j$. Therefore $y$ is the sum of all terms that correspond to antipalindromes. Hence if we substitute $x_{i} = i$ for all $i$, then the value of $y$ would be the sum of the costs of all antipalindromes. So the final answer is:
$$\prod_{j = 1}^{n}(j + (2n + 1 - j)) = \prod_{j = 1}^{n}(2n + 1) = (2n + 1)^{n}.$$

Solution:
Let $n$ be a positive integer. We will find an expression (in terms of $n$) for the sum of the costs of all antipalindromes of length $2n$. Let $\mathcal{P}$ denote the set of all antipalindromes of length $2n$, and let $P$ be an antipalindrome chosen uniformly from $\mathcal{P}$. Note that for each $j = 1,2,\ldots ,n$ the $j^{\mathrm{th}}$ and $(2n + 1 - j)^{\mathrm{th}}$ must be an $A$ and a $B$ in some order. Let $X_{j}$ be the random variable defined by:
$\cdot\ X_{j} = j$ if the $j^{\mathrm{th}}$ letter of $P$ is an $A$ and the $(2n + 1 - j)^{\mathrm{th}}$ letter is a $B$
$\cdot\ X_{j} = 2n + 1 - j$ if the $j^{\mathrm{th}}$ letter of $P$ is a $B$ and the $(2n + 1 - j)^{\mathrm{th}}$ letter is an $A$

Notice that the cost of $P$ is given by the product $X_{1}X_{2}X_{3}\cdot \cdot \cdot X_{n}$. Now consider $f_{j}:\mathcal{P}\to \mathcal{P}$ to be the function which swaps the $j^{\mathrm{th}}$ and $(2n + 1 - j)$ letters of the string. Notice that $f_{j}$ is a bijection that toggles the value of $X_{j}$. This means that $X_{j}$ is equal to $j$ or $(2n + 1 - j)$ with equal probabilities. Therefore
$$\mathbb{P}(X_{j} = j) = \mathbb{P}(X_{j} = 2n + 1 - j) = \frac{1}{2}.$$
Furthermore $f_{j}$ preserves the value of $X_{i}$ for all $i\neq j$. Therefore the variables $X_{i}$ and $X_{j}$ are independent. Therefore the expected value of the cost of $P$ is given by:
$$\mathbb{E}[\mathrm{cost}(P)] = \mathbb{E}\left[\prod_{i = 1}^{n}X_{i}\right] \qquad = \prod_{i = 1}^{n}\mathbb{E}\left[X_{i}\right] \qquad = \prod_{i = 1}^{n}\left(\frac{1}{2} (i) + \frac{1}{2} (2n + 1 - i)\right) \qquad = \prod_{i = 1}^{n}\frac{2n + 1}{2} \qquad = \left(\frac{2n + 1}{2}\right)^{n}.$$
Now the number of antipalindromes of length $2n$ is simply $2^{n}$ (one for each choice of the variables $X_{j}$). Therefore the sum of the costs of all antipalindromes of length $2n$ is simply $2^{n}$ multiplied by the expected value of the cost of $P$. This is
$$2^{n}\times \left(\frac{2n + 1}{2}\right)^{n} = (2n + 1)^{n}.$$