Let D _ k = | array ll 1 2k 2k - 1 n & n ^ 2 + n + 2 & n ^ 2 n & n...

Given that {D}_{k} = \left|$\begin{array}{ll}1 2k 2k - 1 \\ n & {n}^{2} + n + 2 & {n}^{2} \\ n & {n}^{2} + n & {n}^{2} + n + 2\end{array}$\right|

⇒ \mathop{∑}_{k = 1}^{n}{D}_{k} = \left|\begin{cases}∑ 1 & 2 ∑ k & 2 ∑ k - ∑ 1 \\ n & {n}^{2} + n + 2 & {n}^{2} \\ n & {n}^{2} + n & {n}^{2} + n + 2\end{cases}\right|

⇒ \mathop{∑}_{k = 1}^{n}{D}_{k} = \left|\begin{cases}n & 2\dfrac{n\left(n + 1\right)}{2} & 2\left(\dfrac{n\left(n + 1\right)}{2}\right) - n \\ n & {n}^{2} + n + 2 & {n}^{2} \\ n & {n}^{2} + n & {n}^{2} + n + 2\end{cases}\right|

⇒ \mathop{∑}_{k = 1}^{n}{D}_{k} = \left|\begin{cases}n & n\left(n + 1\right) & {n}^{2} \\ n & {n}^{2} + n + 2 & {n}^{2} \\ n & {n}^{2} + n & {n}^{2} + n + 2\end{cases}\right|

${R}_{1} \to {R}_{1} - {R}_{2}$ and ${R}_{2} \to {R}_{2} - {R}_{3}$

⇒ \mathop{∑}_{k = 1}^{n}{D}_{k} = \left|\begin{cases}0 & - 2 & 0 \\ 0 & 2 & - n - 2 \\ n & {n}^{2} + n & {n}^{2} + n + 2\end{cases}\right|

Expand the determinant along ${R}_{1}$.

$⇒ \mathop{∑}_{k = 1}^{n}{D}_{k} = 2\left(\left(- n\right)\left(- n - 2\right)\right) = 96$

$⇒ {n}^{2} + 2n = 48$

$⇒ n = 6, - 8$

So, $n = 6$

Hence this is the required answer.