`|(n,n+1,n+2),(""^(n)P_(n),""^((n+1))P_((n+1)),""^((n+2))P_((n+2))),(""^(n)C_(n),""^((n+1))C_((n+1)),""^((n+2))C_((n+2)))|` is equal to

Options

A. `n(n!)`
B. `(n+1)(n+1)!`
C. `(n+2)(n+2)!`
D. `(n^(2)+n+1)n!`

Answer

To solve the determinant

\[ D = \begin{vmatrix} n & n+1 & n+2 \\ nP_n & (n+1)P_{n+1} & (n+2)P_{n+2} \\ nC_n & (n+1)C_{n+1} & (n+2)C_{n+2} \end{vmatrix} \]

we will follow these steps:

### Step 1: Evaluate the terms in the determinant

1. **Calculate \(nP_n\)**:
\[ nP_n = n! \quad \text{(since } nP_n = \frac{n!}{(n-n)!} = n!\text{)} \]

2. **Calculate \((n+1)P_{n+1}\)**:
\[ (n+1)P_{n+1} = (n+1)! \quad \text{(since } (n+1)P_{n+1} = \frac{(n+1)!}{(n+1-n-1)!} = (n+1)!\text{)} \]

3. **Calculate \((n+2)P_{n+2}\)**:
\[ (n+2)P_{n+2} = (n+2)! \quad \text{(since } (n+2)P_{n+2} = \frac{(n+2)!}{(n+2-n-2)!} = (n+2)!\text{)} \]

4. **Calculate \(nC_n\)**:
\[ nC_n = 1 \quad \text{(since } nC_n = \frac{n!}{n!0!} = 1\text{)} \]

5. **Calculate \((n+1)C_{n+1}\)**:
\[ (n+1)C_{n+1} = 1 \quad \text{(since } (n+1)C_{n+1} = \frac{(n+1)!}{(n+1)!0!} = 1\text{)} \]

6. **Calculate \((n+2)C_{n+2}\)**:
\[ (n+2)C_{n+2} = 1 \quad \text{(since } (n+2)C_{n+2} = \frac{(n+2)!}{(n+2)!0!} = 1\text{)} \]

### Step 2: Substitute values into the determinant

Now substituting these values into the determinant, we have:

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

### Step 3: Simplify the determinant

We can simplify the determinant by performing column operations. Let's perform the following operations:

- \(C_2 \leftarrow C_2 - C_1\)
- \(C_3 \leftarrow C_3 - C_1\)

This gives us:

\[ D = \begin{vmatrix} n & n+1 & n+2 \\ n! & (n+1)! - n! & (n+2)! - n! \\ 1 & 0 & 0 \end{vmatrix} \]

Now, simplifying the second row:

- \((n+1)! - n! = n! \cdot (n+1 - 1) = n!\)
- \((n+2)! - n! = n! \cdot (n+2)(n+1) - n! = n!(n^2 + 3n + 2 - 1) = n!(n^2 + 3n + 1)\)

So we have:

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

### Step 4: Factor out \(n!\)

Factoring out \(n!\) from the second row, we get:

\[ D = n! \cdot \begin{vmatrix} n & n+1 & n+2 \\ 1 & 1 & n^2 + 3n + 1 \\ 1 & 0 & 0 \end{vmatrix} \]

### Step 5: Expand the determinant

Now we can expand this determinant using the first row:

\[ D = n! \left( n \cdot \begin{vmatrix} 1 & n^2 + 3n + 1 \\ 1 & 0 \end{vmatrix} - (n+1) \cdot \begin{vmatrix} 1 & n^2 + 3n + 1 \\ 1 & 0 \end{vmatrix} + (n+2) \cdot \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} \right) \]

Calculating these 2x2 determinants gives us:

- \(\begin{vmatrix} 1 & n^2 + 3n + 1 \\ 1 & 0 \end{vmatrix} = 0 - (n^2 + 3n + 1) = -(n^2 + 3n + 1)\)

- \(\begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} = 0 - 1 = -1\)

### Step 6: Final calculation

Substituting these back into the determinant expression:

\[ D = n! \left( n \cdot (n^2 + 3n + 1) - (n+1) \cdot (n^2 + 3n + 1) - (n+2) \right) \]

This simplifies to:

\[ D = n! \cdot (n^2 + 3n + 1 - (n^2 + 3n + 1) - (n+2)) = n! \cdot (-n - 2) \]

Thus, the final answer is:

\[ D = n! \cdot (-2) \]