If `Delta_a=|(a-1,n,6),((a-1)^2,2n^2,4n-2),((a-1)^3,3n^3,3n^2-3n)|` then `sum_(a=1)^n Delta_a` in equal to

Options

A. 0
B. 1
C. `[(n(n+1))/2][(a(a+1))/2]`
D. none of these

Answer

To solve the problem, we need to evaluate the determinant \(\Delta_a\) given by:

\[ \Delta_a = \begin{vmatrix} a-1 & n & 6 \\ (a-1)^2 & 2n^2 & 4n-2 \\ (a-1)^3 & 3n^3 & 3n^2-3n \end{vmatrix} \]

and then find the summation \(\sum_{a=1}^{n} \Delta_a\).

### Step 1: Calculate the Determinant \(\Delta_a\)

We will use the determinant formula for a \(3 \times 3\) matrix:

\[ \Delta = a(ei - fh) - b(di - fg) + c(dh - eg) \]

where the matrix is:

\[ \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} \]

In our case, we have:

- \(a = a-1\), \(b = n\), \(c = 6\)
- \(d = (a-1)^2\), \(e = 2n^2\), \(f = 4n-2\)
- \(g = (a-1)^3\), \(h = 3n^3\), \(i = 3n^2-3n\)

Now substituting these values into the determinant formula:

\[ \Delta_a = (a-1) \left( 2n^2(3n^2-3n) - (4n-2)(3n^3) \right) - n \left( (a-1)^2(3n^2-3n) - (4n-2)(a-1)^3 \right) + 6 \left( (a-1)^2(3n^3) - 2n^2(a-1)^3 \right) \]

### Step 2: Simplify the Determinant

Now we will simplify each term in the determinant:

1. **First term**:
\[ 2n^2(3n^2 - 3n) = 6n^4 - 6n^3 \]
\[ (4n - 2)(3n^3) = 12n^4 - 6n^3 \]
So,
\[ 2n^2(3n^2 - 3n) - (4n - 2)(3n^3) = (6n^4 - 6n^3) - (12n^4 - 6n^3) = -6n^4 \]

2. **Second term**:
\[ (a-1)^2(3n^2 - 3n) = 3(a-1)^2(n^2 - n) \]
\[ (4n - 2)(a-1)^3 = 4n(a-1)^3 - 2(a-1)^3 \]
So,
\[ (a-1)^2(3n^2 - 3n) - (4n - 2)(a-1)^3 = 3(a-1)^2(n^2 - n) - (4n - 2)(a-1)^3 \]

3. **Third term**:
\[ (a-1)^2(3n^3) - 2n^2(a-1)^3 \]

### Step 3: Combine and Factor

After calculating the determinant, we will combine like terms and factor out common factors. This will yield a polynomial in \(a\).

### Step 4: Find the Summation

Once we have \(\Delta_a\) expressed as a polynomial in \(a\), we can compute the summation:

\[ \sum_{a=1}^{n} \Delta_a \]

This can be done using the formula for the sum of powers of integers.

### Final Result

The final result will be a polynomial expression in \(n\) after evaluating the summation.