1. We need to evaluate $A_r=\begin{vmatrix}r&1&\dfrac{n^2}{2}+\alpha\\[4pt]2r&2&n^2-\beta\\[4pt]3r-2&3&\dfrac{n(3n-1)}{2}\end{vmatrix}$ and then find $2A_{10}-A_8$.

2. Observe that the first two columns are very simple:

• Column 1: $\begin{bmatrix}r\\2r\\3r-2\end{bmatrix}$
• Column 2: $\begin{bmatrix}1\\2\\3\end{bmatrix}$

Use the column operation $C_1 \to C_1-rC_2$ which does not change the determinant.

Then, $C_1=\begin{bmatrix}r\\2r\\3r-2\end{bmatrix}-r\begin{bmatrix}1\\2\\3\end{bmatrix}=\begin{bmatrix}0\\0\\-2\end{bmatrix}.$

So, $A_r=\begin{vmatrix}0&1&\dfrac{n^2}{2}+\alpha\\[4pt]0&2&n^2-\beta\\[4pt]-2&3&\dfrac{n(3n-1)}{2}\end{vmatrix}.$

3. Expand along the first column. Since only the $(3,1)$ entry is nonzero, $A_r=(-2)\cdot(-1)^{3+1}\begin{vmatrix}1&\dfrac{n^2}{2}+\alpha\\[4pt]2&n^2-\beta\end{vmatrix}.$ Because $(-1)^{4}=1$, $A_r=-2\begin{vmatrix}1&\dfrac{n^2}{2}+\alpha\\[4pt]2&n^2-\beta\end{vmatrix}.$

Now compute the $2\times 2$ determinant: $\begin{vmatrix}1&\dfrac{n^2}{2}+\alpha\\[4pt]2&n^2-\beta\end{vmatrix}=1\cdot(n^2-\beta)-2\left(\frac{n^2}{2}+\alpha\right).$

Simplifying, $=n^2-\beta-(n^2+2\alpha)=-\beta-2\alpha.$

Hence, $A_r=-2(-\beta-2\alpha)=2\beta+4\alpha.$

So $A_r$ is actually independent of $r$.

4. Therefore, $A_{10}=4\alpha+2\beta,\qquad A_8=4\alpha+2\beta.$ Thus, $2A_{10}-A_8=2(4\alpha+2\beta)-(4\alpha+2\beta)=4\alpha+2\beta.$

5. Checking options:

• A: $4\alpha+2\beta$ ✅
• B: $0$ ❌
• C: $2n$ ❌
• D: $2\alpha+4\beta$ ❌

So the correct option is A.