Let $$ \Delta_a = \begin{vmatrix} a-1 & n & 6 \\ (a-1)^2 & 2n^2 & 4n - 2 \\ (a-1)^3 & 3n^2 & 3n^2 - 3n \end{vmatrix} $$
My task is to show that $$ \sum_{a = 1}^n \Delta_a = c $$ where c is some constant.
I tried taking a-1 and n common from $C_1$ and $C_2$ respectively and tried to simplify the resulting determinant in the normal method. However, I ended up with a very large result which clearly didn't seem to satisfy the given condition.Please Help.
Recall that the determinant is linear in the columns of the matrix. Since $\Delta_a$ differ only in the first column, we have:
$$\sum_{a=1}^n \Delta_a = \sum_{a=1}^n\begin{vmatrix} a-1 & n & 6 \\ (a-1)^2 & 2n^2 & 4n - 2 \\ (a-1)^3 & 3n^2 & 3n^2 - 3n \end{vmatrix} = \begin{vmatrix} \sum_{a=1}^n (a-1) & n & 6 \\ \sum_{a=1}^n (a-1)^2 & 2n^2 & 4n - 2 \\ \sum_{a=1}^n (a-1)^3 & 3n^2 & 3n^2 - 3n \end{vmatrix}$$
Recognizing the familiar sums $$\sum_{a=1}^n (a-1)= \frac{n(n-1)}2$$$$\sum_{a=1}^n (a-1)^2= \frac{n(n-1)(2n-1)}6$$$$\sum_{a=1}^n (a-1)^3= \left(\frac{n(n-1)}2\right)^2$$
we obtain
$$\begin{vmatrix} \frac{n(n-1)}2 & n & 6 \\ \frac{n(n-1)(2n-1)}6 & 2n^2 & 4n - 2 \\ \left(\frac{n(n-1)}2\right)^2 & 3n^2 & 3n^2 - 3n \end{vmatrix}$$
Now notice that
$$\pmatrix{6 \\ 4n-2 \\ 3n^2-3n} -\frac{12}{n(n-1)} \pmatrix{\frac{n(n-1)}2 \\ \frac{n(n-1)(2n-1)}6 \\ \left(\frac{n(n-1)}2\right)^2} = 0$$
so the columns are linearly dependent. We conclude that the determinant is $0$, which is certainly a constant.