JEE Main 2014 · 19 Apr · Shift 1, Mathematics question 65

Sum the first row over $r=1,\dots,n-1$:

• $\sum r=\frac{n(n-1)}2$;
• $\sum(2r-1)=(n-1)^2$;
• $\sum(3r-2)=\frac{(n-1)(3n-4)}2$.

So $\sum\Delta_r$ is the determinant with this first row. Subtracting $R_3$ from it leaves $(0,0,-4(n-1))$, and the remaining minor is
$$\begin{vmatrix}\frac n2&n-1\\\frac{n(n-1)}2&(n-1)^2\end{vmatrix}=0.$$
The sum is 0, independent of both a and n.