If Δ _ r = | cases r & 2r - 1 & 3r - 2 (n)/(2) & n - 1 & a...

{\Delta}_{r} = \left|\begin{cases}r & \left(2r - 1\right) & \left(3r - 2\right) \\ \dfrac{n}{2} & \left(n - 1\right) & a \\ \dfrac{1}{2}n\left(n - 1\right) & {\left(n - 1\right)}^{2} & \dfrac{1}{2}\left(n - 1\right)\left(3n + 4\right)\end{cases}\right|

Since, the second and the third rows are independent of $r,$ hence the sum is applied to the first row only.

⇒ \mathop{∑}_{r = 1}^{n - 1}{\Delta}_{r} = \left|\begin{cases}\mathop{∑}_{r = 1}^{n - 1}r & 2\mathop{∑}_{r = 1}^{n - 1}r - \mathop{∑}_{r = 1}^{n - 1}1 & 3\mathop{∑}_{r = 1}^{n - 1}r - 2\mathop{∑}_{r = 1}^{n - 1}1 \\ \dfrac{n}{2} & \left(n - 1\right) & a \\ \dfrac{1}{2}n\left(n - 1\right) & {\left(n - 1\right)}^{2} & \dfrac{1}{2}\left(n - 1\right)\left(3n + 4\right)\end{cases}\right|

Using $\mathop{∑}_{r = 1}^{n}r = \dfrac{n\left(n + 1\right)}{2},$ we get

\mathop{∑}_{r = 1}^{n - 1}{\Delta}_{r}=\left|\begin{cases}\dfrac{\left(n - 1\right)n}{2} & \dfrac{2\left(n - 1\right)n}{2} - \left(n - 1\right) & \dfrac{3\left(n - 1\right)n}{2} - 2\left(n - 1\right) \\ \dfrac{n}{2} & \left(n - 1\right) & a \\ \dfrac{1}{2}n\left(n - 1\right) & {\left(n - 1\right)}^{2} & \dfrac{1}{2}\left(n - 1\right)\left(3n + 4\right)\end{cases}\right|

⇒ \mathop{∑}_{r = 1}^{n - 1}{\Delta}_{r}=\left|\begin{cases}\dfrac{1}{2}n\left(n - 1\right) & {\left(n - 1\right)}^{2} & \dfrac{1}{2}\left(n - 1\right)\left(3n + 4\right) \\ \dfrac{n}{2} & \left(n - 1\right) & a \\ \dfrac{1}{2}n\left(n - 1\right) & {\left(n - 1\right)}^{2} & \dfrac{1}{2}\left(n - 1\right)\left(3n + 4\right)\end{cases}\right|

$⇒ \mathop{∑}_{r = 1}^{n - 1}{\Delta}_{r} = 0,$ $( ∵ {R}_{1}$ and ${R}_{3}$ are identical$)$

Hence, the sum is independent of both $n$ and $a .$