Let $$\{a_n\}_{n=0}^{\infty}$$ be a sequence such that $$a_0 = a_1 = 0$$ and $$a_{n+2} = 3a_{n+1} - 2a_n + 1$$, $$\forall n \geq 0$$. Then $$a_{25}a_{23} - 2a_{25}a_{22} - 2a_{23}a_{24} + 4a_{22}a_{24}$$ is equal to

We have the recurrence $$a_{n+2} = 3a_{n+1} - 2a_n + 1$$ with $$a_0 = a_1 = 0$$, and we need to evaluate $$a_{25}a_{23} - 2a_{25}a_{22} - 2a_{23}a_{24} + 4a_{22}a_{24}$$.

We first solve the recurrence. The associated homogeneous equation $$a_{n+2} - 3a_{n+1} + 2a_n = 0$$ has the characteristic equation $$r^2 - 3r + 2 = 0$$, which factors as $$(r-1)(r-2) = 0$$, giving roots $$r = 1$$ and $$r = 2$$. For the particular solution of $$a_{n+2} - 3a_{n+1} + 2a_n = 1$$, since $$r = 1$$ is already a root, we try $$a_n^{(p)} = cn$$. Substituting: $$c(n+2) - 3c(n+1) + 2cn = cn + 2c - 3cn - 3c + 2cn = -c$$. Setting $$-c = 1$$ gives $$c = -1$$.

The general solution is therefore $$a_n = A + B \cdot 2^n - n$$. Applying initial conditions: from $$a_0 = 0$$, we get $$A + B = 0$$; from $$a_1 = 0$$, we get $$A + 2B - 1 = 0$$. Subtracting: $$B = 1$$ and $$A = -1$$. So $$a_n = 2^n - n - 1$$.

We now factor the target expression. Grouping the terms: $$a_{25}a_{23} - 2a_{25}a_{22} - 2a_{23}a_{24} + 4a_{22}a_{24} = a_{25}(a_{23} - 2a_{22}) - 2a_{24}(a_{23} - 2a_{22}) = (a_{25} - 2a_{24})(a_{23} - 2a_{22})$$.

Using $$a_n = 2^n - n - 1$$, we compute $$a_n - 2a_{n-1}$$. We have $$a_{n-1} = 2^{n-1} - (n-1) - 1 = 2^{n-1} - n$$, so $$a_n - 2a_{n-1} = (2^n - n - 1) - 2(2^{n-1} - n) = 2^n - n - 1 - 2^n + 2n = n - 1$$.

Therefore $$a_{25} - 2a_{24} = 24$$ and $$a_{23} - 2a_{22} = 22$$, giving $$(a_{25} - 2a_{24})(a_{23} - 2a_{22}) = 24 \times 22 = 528$$.

Hence, the correct answer is Option B: $$528$$.