the roots of the equations `|{:(.^(x)C_(r),,.^(n-1)C_(r),,.^(n)C_(r)),(.^(x+1)C_(r),,.^(n)C_(r),,.^(n+1)C_(r)),(.^(x+2)C_(r),,.^(n+1)C_(r),,.^(n+2)C_(r)):}|=0`

Options

A. `x=n`
B. `x=n+1`
C. `x=n-1`
D. `x=n-2`

Explanation

To solve the equation given by the determinant:

\[ \begin{vmatrix} {x \choose r} & {n-1 \choose r} & {n \choose r} \\ {x+1 \choose r} & {n \choose r} & {n+1 \choose r} \\ {x+2 \choose r} & {n+1 \choose r} & {n+2 \choose r} \end{vmatrix} = 0 \]

we will follow these steps:

### Step 1: Write the Determinant
We start by rewriting the determinant explicitly:

\[ D = \begin{vmatrix} {x \choose r} & {n-1 \choose r} & {n \choose r} \\ {x+1 \choose r} & {n \choose r} & {n+1 \choose r} \\ {x+2 \choose r} & {n+1 \choose r} & {n+2 \choose r} \end{vmatrix} \]

### Step 2: Use the Binomial Coefficient Formula
Recall that the binomial coefficient can be expressed as:

\[ {x \choose r} = \frac{x!}{r!(x-r)!} \]

Using this formula, we can rewrite each term in the determinant.

### Step 3: Substitute Values
Substituting \(x = n\) into the determinant:

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

### Step 4: Identify Equal Columns
Notice that the first and third columns are equal:

\[ {n \choose r} = {n \choose r} \]

Since two columns are equal, the determinant \(D(n) = 0\).

### Step 5: Substitute \(x = n - 1\)
Now, substitute \(x = n - 1\):

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

Again, the first and second columns are equal:

\[ {n-1 \choose r} = {n-1 \choose r} \]

Thus, \(D(n-1) = 0\).

### Step 6: Conclusion
The roots of the equation are:

\[ x = n \quad \text{and} \quad x = n - 1 \]