If $Δ_r=\begin{vmatrix}r&x&\frac{n (n+1)}{2}\\2r-1&y&n^2\\3r-2&z&\frac{n (3n-1)}{2}\end{vmatrix}$, then $\sum\limits_{r=1}^{n}Δ_r$ is equal to

0

We know that

$\begin{vmatrix}a_1&x&p\\a_2&y&q\\a_3&z&r\end{vmatrix}+\begin{vmatrix}b_2&x&p\\b_2&y&q\\b_3&z&r\end{vmatrix}+\begin{vmatrix}c_2&x&p\\c_2&y&q\\c_3&z&r\end{vmatrix}=\begin{vmatrix}a_1+b_1+c_1&x&p\\a_2+b_2+c_2&y&q\\a_3+b_3+c_3&z&r\end{vmatrix}$

i.e. the sum of the determinants having all columns (or rows) identical except a specific column (or row), say first, is a determinant whose first column (or row) is the sum of the corresponding elements of first columns of various determinants and the remaining columns (or rows) remain same.

Using this property, we have

$\sum\limits_{r=1}^{n}Δ_r=\begin{vmatrix}\sum\limits_{r=1}^{n}r&x&\frac{n (n+1)}{2}\\\sum\limits_{r=1}^{n}(2r-1)&y&n^2\\\sum\limits_{r=1}^{n}(3r-2)&z&\frac{n (3n-1)}{2}\end{vmatrix}$

Now,

$\sum\limits_{r=1}^{n}r=1+2+...+n=\frac{n (n+1)}{2}$

$\sum\limits_{r=1}^{n}(2r-1)=1+3+5+...+(2n-1)=\frac{n}{2}\{1+(2n-1)\}=n^2$

and,

$\sum\limits_{r=1}^{n}(3r-2)=\frac{n}{2}[1+(3n-2)]=\frac{n (3n-1)}{2}$

$∴\sum\limits_{r=1}^{n}Δ_r=\begin{vmatrix}\frac{n (n+1)}{2}&x&\frac{n (n+1)}{2}\\n^2&y&n^2\\\frac{n (3n-1)}{2}&z&\frac{n (3n-1)}{2}\end{vmatrix}=0$   [∵ $C_1$ and $C_3$ are identical]