For alpha, beta in R and a natural number n,l | vedclass

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(A) Given $A_r = \begin{vmatrix} r & 1 & \frac{n^2}{2} + \alpha \ 2r & 2 & n^2 - \beta \ 3r - 2 & 3 & \frac{n(3n - 1)}{2} \end{vmatrix}$.We need to

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$4\alpha + 2\beta$

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(A) Given $A_r = \begin{vmatrix} r & 1 & \frac{n^2}{2} + \alpha \ 2r & 2 & n^2 - \beta \ 3r - 2 & 3 & \frac{n(3n - 1)}{2} \end{vmatrix}$.We need to evaluate $2A_{10} - A_8$.$2A_{10} = \begin{vmatrix} 20 & 1 & \frac{n^2}{2} + \alpha \ 40 & 2 & n^2 - \beta \ 58 & 3 & \frac{n(3n - 1)}{2} \end{vmatrix}$ (Note: $3(10)-2 = 28$,so $2(28)=56$ is incorrect in the prompt,let us re-evaluate the third row: $3(10)-2 = 28$. $2A_{10}$ implies multiplying the first column by $2$,so $2(10)=20, 2(20)=40, 2(28)=56$.)$2A_{10} - A_8 = \begin{vmatrix} 20 & 1 & \frac{n^2}{2} + \alpha \ 40 & 2 & n^2 - \beta \ 56 & 3 & \frac{n(3n - 1)}{2} \end{vmatrix} - \begin{vmatrix} 8 & 1 & \frac{n^2}{2} + \alpha \ 16 & 2 & n^2 - \beta \ 22 & 3 & \frac{n(3n - 1)}{2} \end{vmatrix}$Subtracting the columns: $\begin{vmatrix} 12 & 1 & \frac{n^2}{2} + \alpha \ 24 & 2 & n^2 - \beta \ 34 & 3 & \frac{n(3n - 1)}{2} \end{vmatrix}$Applying $C_1 	o C_1 - 12C_2$: $\begin{vmatrix} 0 & 1 & \frac{n^2}{2} + \alpha \ 0 & 2 & n^2 - \beta \ -2 & 3 & \frac{n(3n - 1)}{2} \end{vmatrix}$Expanding along the first column: $-2 	imes (1 	imes (n^2 - \beta) - 2 	imes (\frac{n^2}{2} + \alpha)) = -2(n^2 - \beta - n^2 - 2\alpha) = -2(-\beta - 2\alpha) = 4\alpha + 2\beta$.

For $\alpha, \beta \in R$ and a natural number $n$,let $A_r = \begin{vmatrix} r & 1 & \frac{n^2}{2} + \alpha \\ 2r & 2 & n^2 - \beta \\ 3r - 2 & 3 & \frac{n(3n - 1)}{2} \end{vmatrix}$. Then $2A_{10} - A_8$ is equal to:

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