Why does the determinant always equal zero for a square matrix of consecutive numbers? This works if the integers are listed in consecutive order either along the rows or columns. Why does the determinant of the square matrix always equal $0$ for $ n > 2 $?

Alternatively, using arithmetic mean, for $3\times 3$ matrix determinant:
$$\begin{vmatrix}
x+1 & x+2 & x+3 \\
x+4 & x+5 & x+6 \\
x+7 & x+8 & x+9 \end{vmatrix}=0,$$
because: $C_1+C_3=2C_2$, which shows linear dependence of the column vectors.
For $4\times 4$ matrix determinant:
$$\begin{vmatrix}
x+1 & x+2 & x+3 & x+4 \\
x+5 & x+6 & x+7 & x+8 \\
x+9 & x+10 & x+11 & x+12 \end{vmatrix}=0, \ \text{because: } 
C_1+C_4=C_2+C_3.$$
In geneal, for $(2n+1)\times (2n+1), n>0,$ matrix determinant:
$$\begin{vmatrix}
x+1&\cdots &x+(n+1) &\cdots &x+(2n+1)\\
x+(2n+1)+1&\cdots&x+(2n+1)+n+1&\cdots&x+2(2n+1)\\
\vdots&\vdots&\vdots &\vdots&\vdots\\
x+2n(2n+1)+1&\cdots &x+2n(2n+1)+n+1&\cdots&x+(2n+1)^2
\end{vmatrix}=0,\\
\text{because: } C_1+C_{2n+1}=C_{n+1}.$$
Can you write the generalization for $(2n)\times (2n),n>0,$ matrix determinant?
