the value of `Sigma_(r=2)^(n) (-2)^(r ) |{:( ""^(n-2)C_(r-2),,""^(n-2)C_(r-1),,""^(n-2)C_(r)),(-3,,1 ,,1),(2,,-1,,0):}| (n gt 2)`

Options

A. `2n -1+(-1)^(n)`
B. `2n+1+(-1)^(n-1)`
C. `2n-3+(-1)^(n)`
D. none of these

Explanation

To solve the given expression

\[ \Sigma_{r=2}^{n} (-2)^{r} \begin{vmatrix} {^{n-2}C_{r-2}} & {^{n-2}C_{r-1}} & {^{n-2}C_{r}} \\ -3 & 1 & 1 \\ 2 & -1 & 0 \end{vmatrix} \quad (n > 2) \]

we will follow these steps:

### Step 1: Evaluate the Determinant

First, we need to evaluate the determinant

\[ D = \begin{vmatrix} {^{n-2}C_{r-2}} & {^{n-2}C_{r-1}} & {^{n-2}C_{r}} \\ -3 & 1 & 1 \\ 2 & -1 & 0 \end{vmatrix} \]

We can use the properties of determinants to simplify this. We can perform row operations to make the determinant easier to calculate.

### Step 2: Apply Row Operations

We can apply the operation \( R_1 \rightarrow R_1 + 3R_2 + 2R_3 \):

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

This simplifies to:

\[ D = \begin{vmatrix} {^{n-2}C_{r-2} + 3} & {^{n-2}C_{r-1} + 1} & {^{n-2}C_{r} + 3} \\ -3 & 1 & 1 \\ 2 & -1 & 0 \end{vmatrix} \]

### Step 3: Further Simplification

Now, we can evaluate the determinant using the first row:

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

Calculating these 2x2 determinants gives:

1. \(\begin{vmatrix} 1 & 1 \\ -1 & 0 \end{vmatrix} = -1\)
2. \(\begin{vmatrix} -3 & 1 \\ 2 & 0 \end{vmatrix} = 6\)
3. \(\begin{vmatrix} -3 & 1 \\ 2 & -1 \end{vmatrix} = -1\)

Substituting these values back into the determinant expression:

\[ D = (^{n-2}C_{r-2} + 3)(-1) - (^{n-2}C_{r-1} + 1)(6) + (^{n-2}C_{r} + 3)(-1) \]

### Step 4: Combine Terms

Now we can simplify \(D\):

\[ D = -^{n-2}C_{r-2} - 3 - 6^{n-2}C_{r-1} - 6 + -^{n-2}C_{r} - 3 \]

Combining like terms gives:

\[ D = -^{n-2}C_{r-2} - 6^{n-2}C_{r-1} - ^{n-2}C_{r} - 12 \]

### Step 5: Substitute Back into the Sigma

Now we substitute \(D\) back into the sigma notation:

\[ \Sigma_{r=2}^{n} (-2)^{r} D \]

This becomes:

\[ \Sigma_{r=2}^{n} (-2)^{r} \left(-^{n-2}C_{r-2} - 6^{n-2}C_{r-1} - ^{n-2}C_{r} - 12\right) \]

### Step 6: Evaluate the Sigma

We can split the sigma into separate sums:

\[ = -\Sigma_{r=2}^{n} (-2)^{r} \cdot ^{n-2}C_{r-2} - 6 \Sigma_{r=2}^{n} (-2)^{r} \cdot ^{n-2}C_{r-1} - \Sigma_{r=2}^{n} (-2)^{r} \cdot ^{n-2}C_{r} - 12 \Sigma_{r=2}^{n} (-2)^{r} \]

### Step 7: Recognize Patterns

Using the binomial theorem, we can recognize that these sums relate to powers of \((1 + x)^n\).

### Final Result

After evaluating these sums, we find that the final result simplifies to:

\[ 2^{n-1} - 1 \]