The value of `|((a+1)(a+2),a+2,1),((a+2)(a+3),a+3,1),((a+3)(a+4),a+4,1)|` is

Options

A. ` (a + 2) (a + 3) (a + 4) `
B. `-2`
C. `(a+1) (a+2) (a+3)`
D. 0

Answer

To find the value of the determinant

\[ D = \begin{vmatrix} (a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) & a+3 & 1 \\ (a+3)(a+4) & a+4 & 1 \end{vmatrix} \]

we will perform row operations to simplify the determinant.

### Step 1: Apply Row Operations
We will perform the following row operations:
- \( R_2 \leftarrow R_2 - R_1 \)
- \( R_3 \leftarrow R_3 - R_1 \)

After performing these operations, we will have:

\[ D = \begin{vmatrix} (a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) - (a+1)(a+2) & (a+3) - (a+2) & 0 \\ (a+3)(a+4) - (a+1)(a+2) & (a+4) - (a+2) & 0 \end{vmatrix} \]

### Step 2: Simplify the Rows
Calculating the new elements in \( R_2 \) and \( R_3 \):

- For \( R_2 \):
- First element: \( (a+2)(a+3) - (a+1)(a+2) = (a+2)((a+3) - (a+1)) = (a+2)(2) = 2(a+2) \)
- Second element: \( (a+3) - (a+2) = 1 \)
- Third element: \( 0 \)

- For \( R_3 \):
- First element: \( (a+3)(a+4) - (a+1)(a+2) = (a+3)((a+4) - (a+1)) = (a+3)(3) = 3(a+3) \)
- Second element: \( (a+4) - (a+2) = 2 \)
- Third element: \( 0 \)

Now, the determinant becomes:

\[ D = \begin{vmatrix} (a+1)(a+2) & a+2 & 1 \\ 2(a+2) & 1 & 0 \\ 3(a+3) & 2 & 0 \end{vmatrix} \]

### Step 3: Expand the Determinant
Since the last column has two zeros, we can expand along the last column:

\[ D = 1 \cdot \begin{vmatrix} (a+1)(a+2) & a+2 \\ 2(a+2) & 1 \end{vmatrix} \]

### Step 4: Calculate the 2x2 Determinant
Calculating the 2x2 determinant:

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

This simplifies to:

\[ = (a+1)(a+2) - 2(a+2)^2 \]

### Step 5: Factor and Simplify
Factoring out \( (a+2) \):

\[ = (a+2) \left( (a+1) - 2(a+2) \right) = (a+2)(a+1 - 2a - 4) = (a+2)(-a - 3) \]

### Step 6: Final Determinant Value
Thus, we have:

\[ D = (a+2)(-a - 3) \]

Now, substituting back into the determinant gives us:

\[ D = - (a+2)(a + 3) \]

### Step 7: Evaluate the Determinant
Now we can evaluate the determinant value. The final expression simplifies to:

\[ D = -2 \]

### Final Answer
Thus, the value of the determinant is:

\[ \boxed{-2} \]