Question

If $\Delta_{r} = \begin{vmatrix} r-1 & n & 6 \\ (r-1)^{2} & 2n^{2} & 4n-2 \\ (r-1)^{3} & 3n^{3} & 3n^{2} - 3n \end{vmatrix}$, then prove that $\sum_{r=1}^{n} \Delta_{r}$ is equal to 0.

Answer 1

The statement $\sum_{r=1}^{n} \Delta_{r} = 0$ is proved above.

Answer 2

The determinant $\Delta_r$ is expressed as a polynomial in $x = r-1$. By expanding the determinant along the first column, we find $\Delta_r = A x + B x^2 + C x^3$, where $A, B, C$ are coefficients depending on $n$. The sum $\sum_{r=1}^{n} \Delta_r$ becomes $\sum_{x=0}^{n-1} (A x + B x^2 + C x^3)$. Using formulas for sums of powers $\sum x$, $\sum x^2$, $\sum x^3$, and substituting the specific values of $A, B, C$ derived from the determinant, the entire sum simplifies to $0$.