Let $D_k=\left|\begin{array}{c}\begin{array}{ccc}1& 2k& 2k-1\\ n& n^2+n+2& n^2\\ n& n^2+n& n^2+n+2\end{array}\end{array}\right|$. If $\sum _{k=1}^n{D}_k=96$, then $n$ is equal to _________.

Given that $D_k=\left|\begin{array}{c}\begin{array}{ccc}1& 2k& 2k-1\\ n& n^2+n+2& n^2\\ n& n^2+n& n^2+n+2\end{array}\end{array}\right|$

$\Rightarrow \sum _{k=1}^n{D}_k=\left|\begin{array}{ccc}\sum 1& 2\sum k& 2\sum k-\sum 1\\ n& n^2+n+2& n^2\\ n& n^2+n& n^2+n+2\end{array}\right|$

$\Rightarrow \sum _{k=1}^n{D}_k=\left|\begin{array}{ccc}n& 2\frac{n\left(n+1\right)}{2}& 2\left(\frac{n\left(n+1\right)}{2}\right)-n\\ n& n^2+n+2& n^2\\ n& n^2+n& n^2+n+2\end{array}\right|$

$\Rightarrow \sum _{k=1}^n{D}_k=\left|\begin{array}{ccc}n& n\left(n+1\right)& n^2\\ n& n^2+n+2& n^2\\ n& n^2+n& n^2+n+2\end{array}\right|$

$R_1\to R_1-R_2$ and $R_2\to R_2-R_3$

$\Rightarrow \sum _{k=1}^n{D}_k=\left|\begin{array}{ccc}0& -2& 0\\ 0& 2& -n-2\\ n& n^2+n& n^2+n+2\end{array}\right|$

Expand the determinant along $R_1$.

$\Rightarrow \sum _{k=1}^n{D}_k=2\left(\left(-n\right)\left(-n-2\right)\right)=96$

$\Rightarrow n^2+2n=48$

$\Rightarrow n=6, -8$

So, $n=6$

Hence this is the required answer.