For α, β ∈ mathbbR and a natural number n, let A_r= | … |. Then

$$A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 2 r & 2 & n^2-\beta \\ 3 r-2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|$$Ar​=​r2r3r−2​123​2n2​+αn2−β2n(3n−1)​​​

$2 \mathrm{~A}_{10}-\mathrm{A}_8=\left|\begin{array}{ccc}20 & 1 & \frac{\mathrm{n}^2}{2}+\alpha \\ 40 & 2 & \mathrm{n}^2-\beta \\ 56 & 3 & \frac{\mathrm{n}(3 \mathrm{n}-1)}{2}\end{array}\right|-\left|\begin{array}{ccc}8 & 1 & \frac{\mathrm{n}^2}{2}+\alpha \\ 16 & 2 & \mathrm{n}^2-\beta \\ 22 & 3 & \frac{\mathrm{n}(3 \mathrm{n}-1)}{2}\end{array}\right|$2 A10​−A8​=​204056​123​2n2​+αn2−β2n(3n−1)​​​−​81622​123​2n2​+αn2−β2n(3n−1)​​​ $\begin{aligned} & \Rightarrow\left|\begin{array}{ccc}12 & 1 & \frac{n^2}{2}+\alpha \\ 24 & 2 & n^2-\beta \\ 34 & 3 & \frac{n(3 n-1)}{2}\end{array}\right| \\ & \Rightarrow\left|\begin{array}{ccc}0 & 1 & \frac{n^2}{2}+\alpha \\ 0 & 2 & n^2-\beta \\ -2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right| \\ & \Rightarrow-2\left(\left(n^2-\beta\right)-\left(n^2+2 \alpha\right)\right) \\ & \Rightarrow-2(-\beta-2 \alpha) \Rightarrow 4 \alpha+2 \beta\end{aligned}$​⇒​122434​123​2n2​+αn2−β2n(3n−1)​​​⇒​00−2​123​2n2​+αn2−β2n(3n−1)​​​⇒−2((n2−β)−(n2+2α))⇒−2(−β−2α)⇒4α+2β​