For $$\alpha, \beta \in \mathbb{R}$$ and a natural number $$n$$, let $$A_r = \begin{vmatrix} r & 1 & \frac{n^2}{2} + \alpha \\ 2r & 2 & n^2 - \beta \\ 3r - 2 & 3 & \frac{n(3n-1)}{2} \end{vmatrix}$$. Then $$\sum_{r=1}^{n} A_r$$ is

Since the determinant is linear in Column 1, we can bring the sum inside:

$$\sum_{r=1}^n A_r = \begin{vmatrix} \sum r & 1 & \frac{n^2}{2} + \alpha \\ \sum 2r & 2 & n^2 - \beta \\ \sum (3r - 2) & 3 & \frac{n(3n-1)}{2} \end{vmatrix}$$

Using standard AP sum formulas for $$r = 1 \text{ to } n$$:

$$\sum r = \frac{n^2+n}{2}, \quad \sum 2r = n^2+n, \quad \sum (3r-2) = \frac{n(3n-1)}{2}$$

Substitute $$n = 1$$ as a shortcut:

$$\det = \begin{vmatrix} 1 & 1 & \frac{1}{2}+\alpha \\ 2 & 2 & 1-\beta \\ 1 & 3 & 1 \end{vmatrix}$$

Apply row operation $$R_2 \to R_2 - 2R_1$$:

$$\det = \begin{vmatrix} 1 & 1 & \frac{1}{2}+\alpha \\ 0 & 0 & -2\alpha-\beta \\ 1 & 3 & 1 \end{vmatrix} = -(-2\alpha - \beta)(3 - 1) = 4\alpha + 2\beta$$