For `alpha, beta in R` and a natural number n, let `A_r = |[r, 1, (n^2)/2 + alpha],[2r, 2, n^2 – beta],[3r – 2, 3, frac{n(3n - 1)}{2}]|`. Then `2A_(10) – A_8` is

Options

A. 0
B. `4 alpha + 2 beta`
C. 2n
D. `2 alpha + 4 beta`

Answer

To solve the problem, we need to evaluate the determinant \( A_r \) given by:

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

We are tasked with finding \( 2A_{10} - A_8 \).

### Step 1: Factor out common terms
First, we can factor out common terms from the rows. Notice that we can factor out 2 from the second row:

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

### Step 2: Row operation
Next, we perform a row operation \( R_1 \leftarrow R_1 - R_2 \):

\[ A_r = 2 \begin{vmatrix} 0 & 0 & \frac{n^2}{2} + \alpha - \frac{n^2 - \beta}{2} \\ r & 1 & \frac{n^2 - \beta}{2} \\ 3r - 2 & 3 & \frac{n(3n - 1)}{2} \end{vmatrix} \]

This simplifies to:

\[ A_r = 2 \begin{vmatrix} 0 & 0 & \alpha + \frac{\beta}{2} \\ r & 1 & \frac{n^2 - \beta}{2} \\ 3r - 2 & 3 & \frac{n(3n - 1)}{2} \end{vmatrix} \]

### Step 3: Evaluate the determinant
Since the first column has zeros, we can expand the determinant along the first column:

\[ A_r = 2 \left( \alpha + \frac{\beta}{2} \right) \begin{vmatrix} r & 1 \\ 3r - 2 & 3 \end{vmatrix} \]

Calculating the 2x2 determinant:

\[ \begin{vmatrix} r & 1 \\ 3r - 2 & 3 \end{vmatrix} = r \cdot 3 - 1 \cdot (3r - 2) = 3r - (3r - 2) = 2 \]

Thus,

\[ A_r = 2 \left( \alpha + \frac{\beta}{2} \right) \cdot 2 = 4 \left( \alpha + \frac{\beta}{2} \right) = 4\alpha + 2\beta \]

### Step 4: Calculate \( A_{10} \) and \( A_8 \)
Now we can find \( A_{10} \) and \( A_8 \):

\[ A_{10} = 4\alpha + 2\beta \]
\[ A_8 = 4\alpha + 2\beta \]

### Step 5: Calculate \( 2A_{10} - A_8 \)
Now substituting these values into the expression \( 2A_{10} - A_8 \):

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

### Final Answer
Thus, the final answer is:

\[ \boxed{4\alpha + 2\beta} \]