1. We are given

$$D_k=\begin{vmatrix} 1 & 2k & 2k-1\\ n & n^2+n+2 & n^2\\ n & n^2+n & n^2+n+2 \end{vmatrix}$$

and

$$\sum_{k=1}^{n} D_k = 96.$$

We must find $n$.

2. Simplify the determinant.

Apply row operation: $R_3 \to R_3 - R_2$ Then

$$D_k= \begin{vmatrix} 1 & 2k & 2k-1\\ n & n^2+n+2 & n^2\\ 0 & -2 & n+2 \end{vmatrix}.$$

3. Expand along the third row:

$$D_k = 0\cdot C_{31} + (-2)C_{32} + (n+2)C_{33}.$$

It is easier to directly compute:

$$D_k = 1\begin{vmatrix} n^2+n+2 & n^2 \\ -2 & n+2 \end{vmatrix} -2k\begin{vmatrix} n & n^2 \\ 0 & n+2 \end{vmatrix} +(2k-1)\begin{vmatrix} n & n^2+n+2 \\ 0 & -2 \end{vmatrix}.$$

Now evaluate each minor:

• First minor:

$$\begin{vmatrix} n^2+n+2 & n^2 \\ -2 & n+2 \end{vmatrix} =(n^2+n+2)(n+2)-n^2(-2).$$

Expand:

$$(n^2+n+2)(n+2)=n^3+3n^2+4n+4,$$

so

$$= n^3+3n^2+4n+4+2n^2 = n^3+5n^2+4n+4.$$

• Second minor:

$$\begin{vmatrix} n & n^2 \\ 0 & n+2 \end{vmatrix}=n(n+2)=n^2+2n.$$

Hence contribution is

$$-2k(n^2+2n).$$

• Third minor:

$$\begin{vmatrix} n & n^2+n+2 \\ 0 & -2 \end{vmatrix}=n(-2)=-2n.$$

Hence contribution is

$$(2k-1)(-2n)=-4kn+2n.$$

Therefore,

$$D_k = n^3+5n^2+4n+4 -2k(n^2+2n) -4kn +2n.$$

Combine terms:

$$D_k = n^3+5n^2+6n+4 -2kn^2-8kn.$$

So

$$D_k = n^3+5n^2+6n+4 -2k(n^2+4n).$$

4. Sum from $k=1$ to $n$:

$$\sum_{k=1}^n D_k = \sum_{k=1}^n \left(n^3+5n^2+6n+4\right) -2(n^2+4n)\sum_{k=1}^n k.$$

Since the first part is constant in $k$,

$$\sum_{k=1}^n D_k = n(n^3+5n^2+6n+4)-2(n^2+4n)\cdot \frac{n(n+1)}{2}.$$

This becomes

$$= n(n^3+5n^2+6n+4) - (n^2+4n)n(n+1).$$

Factor $n$:

$$= n\left[(n^3+5n^2+6n+4) - (n^2+4n)(n+1)\right].$$

Now,

$$(n^2+4n)(n+1)=n^3+5n^2+4n.$$

So,

$$\sum_{k=1}^n D_k = n\left[(n^3+5n^2+6n+4)-(n^3+5n^2+4n)\right] = n(2n+4).$$

Thus,

$$\sum_{k=1}^n D_k = 2n^2+4n.$$

5. Given this equals $96$:

$$2n^2+4n=96.$$

Divide by $2$:

$$n^2+2n=48.$$

So,

$$n^2+2n-48=0.$$

Factor:

$$(n+8)(n-6)=0.$$

Hence,

$$n=6 \quad \text{or} \quad n=-8.$$

Since the sum is from $k=1$ to $n$, $n$ must be a positive integer. Therefore,

$$\boxed{n=6}.$$

6. Comparison with stored answer: Stored correct answer = $6$. Our derived answer also is $6$, so they agree.