If Δ _ =| … |, then the value of Σ _r=1^n-1 Δ _

$${\Delta }_{r}=|\begin{matrix}r & (2r-1) & (3r-2) \\ \frac{n}{2} & (n-1) & a \\ \frac{1}{2}n(n-1) & {(n-1)}^{2} & \frac{1}{2}(n-1)(3n+4)\end{matrix}|$$Δr​=∣r2n​21​n(n−1)​(2r−1)(n−1)(n−1)2​(3r−2)a21​(n−1)(3n+4)​∣

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

$$\Rightarrow \sum _{r=1}^{n-1}{\Delta }_{r}=|\begin{matrix}\sum _{r=1}^{n-1}r & 2\sum _{r=1}^{n-1}r-\sum _{r=1}^{n-1}1 & 3\sum _{r=1}^{n-1}r-2\sum _{r=1}^{n-1}1 \\ \frac{n}{2} & (n-1) & a \\ \frac{1}{2}n(n-1) & {(n-1)}^{2} & \frac{1}{2}(n-1)(3n+4)\end{matrix}|$$⇒r=1∑n−1​Δr​=∣∑r=1n−1​r2n​21​n(n−1)​2∑r=1n−1​r−∑r=1n−1​1(n−1)(n−1)2​3∑r=1n−1​r−2∑r=1n−1​1a21​(n−1)(3n+4)​∣

Using $\displaystyle \sum _{r=1}^{n}r=\frac{n(n+1)}{2},$r=1∑n​r=2n(n+1)​, we get

$$\sum _{r=1}^{n-1}{\Delta }_{r}=|\begin{matrix}\frac{(n-1)n}{2} & \frac{2(n-1)n}{2}-(n-1) & \frac{3(n-1)n}{2}-2(n-1) \\ \frac{n}{2} & (n-1) & a \\ \frac{1}{2}n(n-1) & {(n-1)}^{2} & \frac{1}{2}(n-1)(3n+4)\end{matrix}|$$r=1∑n−1​Δr​=∣2(n−1)n​2n​21​n(n−1)​22(n−1)n​−(n−1)(n−1)(n−1)2​23(n−1)n​−2(n−1)a21​(n−1)(3n+4)​∣

$$\Rightarrow \sum _{r=1}^{n-1}{\Delta }_{r}=|\begin{matrix}\frac{1}{2}n(n-1) & {(n-1)}^{2} & \frac{1}{2}(n-1)(3n+4) \\ \frac{n}{2} & (n-1) & a \\ \frac{1}{2}n(n-1) & {(n-1)}^{2} & \frac{1}{2}(n-1)(3n+4)\end{matrix}|$$⇒r=1∑n−1​Δr​=∣21​n(n−1)2n​21​n(n−1)​(n−1)2(n−1)(n−1)2​21​(n−1)(3n+4)a21​(n−1)(3n+4)​∣

$\displaystyle \Rightarrow \sum _{r=1}^{n-1}{\Delta }_{r}=0,$⇒r=1∑n−1​Δr​=0, $(\because {R}_{1}$(∵R1​ and ${R}_{3}$R3​ are identical$)$)

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