For $\alpha, \beta \in \mathbb{R}$ and a natural number $n$, let $A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \ 2 r & 2 & n^2-\beta \ 3 r-2 & 3 & \frac{n(3 n-1)}{2}\end{array} ight|$. Then

Question

For $\alpha, \beta \in \mathbb{R}$ and a natural number $n$, let $A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \ 2 r & 2 & n^2-\beta \ 3 r-2 & 3 & \frac{n(3 n-1)}{2}\end{array}ight|$. Then

Aicharya · Accepted Answer

$A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 2 r & 2 & n^2-\beta \\ 3 r-2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|$
$2 \mathrm{~A}_{10}-\mathrm{A}_8=\left|\begin{array}{ccc}20 & 1 & \frac{\mathrm{n}^2}{2}+\alpha \\ 40 & 2 & \mathrm{n}^2-\beta \\ 56 & 3 & \frac{\mathrm{n}(3 \mathrm{n}-1)}{2}\end{array}\right|-\left|\begin{array}{ccc}8 & 1 & \frac{\mathrm{n}^2}{2}+\alpha \\ 16 & 2 & \mathrm{n}^2-\beta \\ 22 & 3 & \frac{\mathrm{n}(3 \mathrm{n}-1)}{2}\end{array}\right|$ $\begin{aligned} & \Rightarrow\left|\begin{array}{ccc}12 & 1 & \frac{n^2}{2}+\alpha \\ 24 & 2 & n^2-\beta \\ 34 & 3 & \frac{n(3 n-1)}{2}\end{array}\right| \\ & \Rightarrow\left|\begin{array}{ccc}0 & 1 & \frac{n^2}{2}+\alpha \\ 0 & 2 & n^2-\beta \\ -2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right| \\ & \Rightarrow-2\left(\left(n^2-\beta\right)-\left(n^2+2 \alpha\right)\right) \\ & \Rightarrow-2(-\beta-2 \alpha) \Rightarrow 4 \alpha+2 \beta\end{aligned}$

For $\alpha, \beta \in \mathbb{R}$ and a natural number $n$, let $A_r=\left|\begin{array}{ccc}r & 1 &…

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