The value of the determinant `|(1,1,1),(.^(m)C_(1),.^(m +1)C_(1),.^(m+2)C_(1)),(.^(m)C_(2),.^(m +1)C_(2),.^(m+2)C_(2))|` is equal to

Options

A. 1
B. -1
C. 0
D. none of these

Explanation

To find the value of the determinant

\[ D = \begin{vmatrix} 1 & 1 & 1 \\ \binom{m}{1} & \binom{m+1}{1} & \binom{m+2}{1} \\ \binom{m}{2} & \binom{m+1}{2} & \binom{m+2}{2} \end{vmatrix} \]

we will simplify it step by step.

### Step 1: Write the determinant

We start with the determinant as given:

\[ D = \begin{vmatrix} 1 & 1 & 1 \\ \binom{m}{1} & \binom{m+1}{1} & \binom{m+2}{1} \\ \binom{m}{2} & \binom{m+1}{2} & \binom{m+2}{2} \end{vmatrix} \]

### Step 2: Simplify the second and third rows

Recall the binomial coefficient properties:

- \(\binom{n}{1} = n\)
- \(\binom{n}{2} = \frac{n(n-1)}{2}\)

Thus, we can rewrite the second and third rows:

\[ D = \begin{vmatrix} 1 & 1 & 1 \\ m & m + 1 & m + 2 \\ \frac{m(m-1)}{2} & \frac{(m+1)m}{2} & \frac{(m+2)(m+1)}{2} \end{vmatrix} \]

### Step 3: Perform column operations

To simplify the determinant, we can perform column operations. We will subtract the first column from the second and third columns:

\[ D = \begin{vmatrix} 1 & 1 & 1 \\ m & (m + 1 - m) & (m + 2 - m) \\ \frac{m(m-1)}{2} & \frac{(m+1)m}{2} - \frac{m(m-1)}{2} & \frac{(m+2)(m+1)}{2} - \frac{m(m-1)}{2} \end{vmatrix} \]

This simplifies to:

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

### Step 4: Simplify the second row

The second row simplifies to:

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

### Step 5: Calculate the determinant

Now we can expand the determinant along the first row:

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

Calculating this determinant results in:

\[ D = 1 \cdot \left(1 \cdot \frac{(m+2)(m+1) - m(m-1)}{2} - 2 \cdot \frac{(m+1)m - m(m-1)}{2}\right) \]

### Step 6: Final simplification

After simplification, we find that the determinant evaluates to 1.

Thus, the value of the determinant is:

\[ \boxed{1} \]