For alpha, beta in mathrm{R} and a natural nu | vedclass

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Question

$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}\

Vedclass · Accepted Answer

$4 \alpha+2 \beta$

Vedclass · Answer

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

For $\alpha, \beta \in \mathrm{R}$ and a natural number $\mathrm{n}$, let

$A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 2 r & 2 & n^2-\beta \\ 3 r-2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|$. Then $2 A_{10}-A_8$

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For $\alpha, \beta \in \mathrm{R}$ and a natural number $\mathrm{n}$, let $A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 2 r & 2 & n^2-\beta \\ 3 r-2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|$. Then $2 A_{10}-A_8$