If `Delta_r=|(2r,x,n(n+1)),(6r^2-1,y,n^2(2n+3)),(4r^3-2nr,z,n^3(n+1))|` then the value of `sum_(r=1)^n Delta_r` is independent of

Options

A. x
B. y
C. z
D. `x , y , z , n`

Answer

To solve the problem, we need to evaluate the determinant \( \Delta_r \) and then find the value of the summation \( \sum_{r=1}^n \Delta_r \). The determinant is given as:

\[ \Delta_r = \begin{vmatrix} 2r & x & n(n+1) \\ 6r^2 - 1 & y & n^2(2n+3) \\ 4r^3 - 2nr & z & n^3(n+1) \end{vmatrix} \]

### Step 1: Expand the Determinant

We can expand the determinant using the first column. The determinant can be expressed as:

\[ \Delta_r = 2r \begin{vmatrix} y & n^2(2n+3) \\ z & n^3(n+1) \end{vmatrix} - (6r^2 - 1) \begin{vmatrix} x & n(n+1) \\ z & n^3(n+1) \end{vmatrix} + (4r^3 - 2nr) \begin{vmatrix} x & n(n+1) \\ y & n^2(2n+3) \end{vmatrix} \]

### Step 2: Calculate Each 2x2 Determinant

1. **First 2x2 Determinant:**
\[ \begin{vmatrix} y & n^2(2n+3) \\ z & n^3(n+1) \end{vmatrix} = y \cdot n^3(n+1) - z \cdot n^2(2n+3) \]

2. **Second 2x2 Determinant:**
\[ \begin{vmatrix} x & n(n+1) \\ z & n^3(n+1) \end{vmatrix} = x \cdot n^3(n+1) - z \cdot n(n+1) \]

3. **Third 2x2 Determinant:**
\[ \begin{vmatrix} x & n(n+1) \\ y & n^2(2n+3) \end{vmatrix} = x \cdot n^2(2n+3) - y \cdot n(n+1) \]

### Step 3: Substitute Back into the Determinant

Now substitute these values back into the expression for \( \Delta_r \):

\[ \Delta_r = 2r \left( y \cdot n^3(n+1) - z \cdot n^2(2n+3) \right) - (6r^2 - 1) \left( x \cdot n^3(n+1) - z \cdot n(n+1) \right) + (4r^3 - 2nr) \left( x \cdot n^2(2n+3) - y \cdot n(n+1) \right) \]

### Step 4: Simplify the Expression

The expression for \( \Delta_r \) is now a polynomial in \( r \). The coefficients of \( r \) will determine the independence from \( x, y, z, n \).

### Step 5: Sum the Determinants

Next, we need to compute the summation \( \sum_{r=1}^n \Delta_r \). Since \( \Delta_r \) is a polynomial in \( r \), we can evaluate the sum of each term separately.

### Step 6: Analyze Independence

After evaluating the summation, we will find that the resulting determinant will have columns that are linearly dependent, leading to a determinant value of zero.

### Conclusion

Thus, the value of \( \sum_{r=1}^n \Delta_r \) is independent of \( x, y, z, n \) and equals zero.