proving the sum of a few determinant values as a constant 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.
 A: 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.
