"For $\alpha, \beta \in \mathbb{R}$ and a natural number $n$, let $$A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 2 r & 2 & n^2-\beta \\ 3 r-2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|$$. Then $2 A_{10}-A_8$ is"

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We are given a determinant $A_r$ defined as:

$$A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 2 r & 2 & n^2-\beta \\ 3 r-2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|$$

We need to find the value of $2 A_{10} - A_8$.

Let's calculate the determinant $A_r$. We can use properties of determinants or expand it directly. A helpful trick is to use row operations to simplify the determinant before expanding. Let's perform the following row operations:

Replace Row 2 with (Row 2) - 2 $\times$ (Row 1)

Replace Row 3 with (Row 3) - 3 $\times$ (Row 1)

The determinant becomes:

$$A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 2 r - 2r & 2 - 2(1) & (n^2-\beta) - 2(\frac{n^2}{2}+\alpha) \\ (3 r-2) - 3r & 3 - 3(1) & \frac{n(3 n-1)}{2} - 3(\frac{n^2}{2}+\alpha)\end{array}\right|$$

Let's simplify the entries in the new rows:

Row 2:

First element: $2r - 2r = 0$

Second element: $2 - 2 = 0$

Third element: $(n^2-\beta) - 2(\frac{n^2}{2}+\alpha) = n^2 - \beta - n^2 - 2\alpha = -2\alpha - \beta$

Row 3:

First element: $(3r-2) - 3r = -2$

Second element: $3 - 3 = 0$

Third element: $\frac{n(3 n-1)}{2} - 3(\frac{n^2}{2}+\alpha) = \frac{3n^2 - n}{2} - \frac{3n^2}{2} - 3\alpha = \frac{3n^2 - n - 3n^2}{2} - 3\alpha = -\frac{n}{2} - 3\alpha$

So the determinant simplifies to:

$$A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 0 & 0 & -2\alpha - \beta \\ -2 & 0 & -\frac{n}{2} - 3\alpha\end{array}\right|$$

Now, we can easily calculate this determinant by expanding along the second column, since it has two zeros. The expansion along the second column is:

$$A_r = -1 \cdot \left|\begin{array}{cc} 0 & -2\alpha - \beta \\ -2 & -\frac{n}{2} - 3\alpha \end{array}\right|$$

$$A_r = -1 \cdot \left( (0) \times (-\frac{n}{2} - 3\alpha) - (-2) \times (-2\alpha - \beta) \right)$$

$A_r = -1 \cdot \left( 0 - (4\alpha + 2\beta) \right)$

$A_r = -1 \cdot (-4\alpha - 2\beta)$

$A_r = 4\alpha + 2\beta$

Notice that the value of the determinant $A_r$ does not depend on $r$ at all! It's a constant value determined by $\alpha$ and $\beta$.

Now we need to find the value of $2 A_{10} - A_8$.

Since $A_r = 4\alpha + 2\beta$ for any value of $r$, we have:

$A_{10} = 4\alpha + 2\beta$

$A_8 = 4\alpha + 2\beta$

So, $2 A_{10} - A_8 = 2 (4\alpha + 2\beta) - (4\alpha + 2\beta)$

$2 A_{10} - A_8 = 8\alpha + 4\beta - 4\alpha - 2\beta$

$2 A_{10} - A_8 = (8\alpha - 4\alpha) + (4\beta - 2\beta)$

$2 A_{10} - A_8 = 4\alpha + 2\beta$

The value of the expression $2 A_{10} - A_8$ is $4\alpha + 2\beta$.

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For $\alpha, \beta \in \mathbb{R}$ and a natural number $n$, let $$A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 2 r & 2 & n^2-\beta \\ 3 r-2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|$$. Then $2 A_{10}-A_8$ is

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For $\alpha, \beta \in \mathbb{R}$ and a natural number $n$, let $$A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 2 r & 2 & n^2-\beta \\ 3 r-2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|$$. Then $2 A_{10}-A_8$ is

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