Key Concept: Summation Property of Determinants

The foundation of solving this problem lies in a crucial property of determinants: If the elements of a single row or a single column of a determinant are expressed as a sum of functions of an index $k$, while the elements of the other rows/columns are independent of $k$, then the sum of such determinants can be written as a single determinant where the elements of that specific row/column are replaced by their respective sums. Mathematically, if $D_k = \left|\begin{array}{ccc} f_1(k) & f_2(k) & f_3(k) \\ a & b & c \\ d & e & f \end{array}\right|$, then $\sum_{k=1}^n D_k = \left|\begin{array}{ccc} \sum_{k=1}^n f_1(k) & \sum_{k=1}^n f_2(k) & \sum_{k=1}^n f_3(k) \\ a & b & c \\ d & e & f \end{array}\right|$.

We will also use the standard summation formulas:

1. $\sum_{k=1}^n 1 = n$
2. $\sum_{k=1}^n k = \frac{n(n+1)}{2}$

---

Step-by-Step Solution

1. Apply the Summation Property to the Determinant

We are given the determinant $D_k=\left|\begin{array}{ccc}1 & 2 k & 2 k-1 \\ n & n^{2}+n+2 & n^{2} \\ n & n^{2}+n & n^{2}+n+2\end{array}\right|$. Notice that only the elements in the first row depend on $k$. The elements in the second and third rows are independent of $k$. This allows us to apply the summation property directly.

We need to calculate $\sum_{k=1}^{n} D_k$: $\sum_{k=1}^{n} D_k = \left|\begin{array}{ccc} \sum_{k=1}^{n} 1 & \sum_{k=1}^{n} (2k) & \sum_{k=1}^{n} (2k-1) \\ n & n^{2}+n+2 & n^{2} \\ n & n^{2}+n & n^{2}+n+2 \end{array}\right|$

Now, let's evaluate each sum in the first row:

• For the first element: $\sum_{k=1}^{n} 1 = n$.
• For the second element: $\sum_{k=1}^{n} (2k) = 2 \sum_{k=1}^{n} k = 2 \cdot \frac{n(n+1)}{2} = n(n+1)$.
• For the third element: $\sum_{k=1}^{n} (2k-1) = 2 \sum_{k=1}^{n} k - \sum_{k=1}^{n} 1 = 2 \cdot \frac{n(n+1)}{2} - n = n(n+1) - n = n^2+n-n = n^2$.

Substituting these sums back into the determinant, we get: $\sum_{k=1}^{n} D_k = \left|\begin{array}{ccc} n & n(n+1) & n^2 \\ n & n^2+n+2 & n^2 \\ n & n^2+n & n^2+n+2 \end{array}\right|$ This is the first crucial step.

2. Simplify the Determinant using Row Operations

Our goal now is to evaluate this $3 \times 3$ determinant. A common strategy to simplify determinants is to create as many zeros as possible in a row or column using elementary row/column operations. This makes the expansion of the determinant much easier.

Observe the first column: all elements are $n$. This is a strong hint to perform row operations involving $R_1, R_2, R_3$ to create zeros in the first column.

• Operation 1: Apply $R_1 \leftarrow R_1 - R_2$.

  • This operation aims to make the first element of the first row zero.
  • New $R_1$ elements:
    • $n - n = 0$
    • $n(n+1) - (n^2+n+2) = (n^2+n) - (n^2+n+2) = -2$
    • $n^2 - n^2 = 0$
  • The determinant becomes: $\left|\begin{array}{ccc} 0 & -2 & 0 \\ n & n^2+n+2 & n^2 \\ n & n^2+n & n^2+n+2 \end{array}\right|$

• Operation 2: Apply $R_2 \leftarrow R_2 - R_3$.

  • This operation aims to make the first element of the second row zero.
  • New $R_2$ elements:
    • $n - n = 0$
    • $(n^2+n+2) - (n^2+n) = 2$
    • $n^2 - (n^2+n+2) = -n-2$
  • The determinant now simplifies to: $\left|\begin{array}{ccc} 0 & -2 & 0 \\ 0 & 2 & -n-2 \\ n & n^2+n & n^2+n+2 \end{array}\right|$
  • Tip: Row/column operations ($R_i \leftarrow R_i + cR_j$ or $C_i \leftarrow C_i + cC_j$) do not change the value of the determinant. This is why we can perform them freely to simplify.

3. Expand the Determinant

Now that we have two zeros in the first column, expanding the determinant along the first column ($C_1$) is the most efficient way. The expansion is $0 \cdot C_{11} + 0 \cdot C_{21} + n \cdot C_{31}$, where $C_{ij}$ is the cofactor of the element at row $i$, column $j$.

$\left|\begin{array}{ccc} 0 & -2 & 0 \\ 0 & 2 & -n-2 \\ n & n^2+n & n^2+n+2 \end{array}\right| = n \cdot (-1)^{3+1} \cdot \left|\begin{array}{cc} -2 & 0 \\ 2 & -n-2 \end{array}\right|$ (Note: $(-1)^{3+1}$ is for the element $n$ in the third row, first column)

Now, calculate the $2 \times 2$ determinant: $\left|\begin{array}{cc} -2 & 0 \\ 2 & -n-2 \end{array}\right| = (-2)(-n-2) - (0)(2) = 2(n+2) - 0 = 2(n+2)$

So, the value of the determinant is $n \cdot 1 \cdot 2(n+2) = 2n(n+2)$.

4. Solve the Equation for $n$

We are given that $\sum_{k=1}^{n} D_k = 96$. Substituting our calculated determinant value: $2n(n+2) = 96$ Divide both sides by 2: $n(n+2) = 48$ Expand the left side: $n^2 + 2n = 48$ Rearrange into a standard quadratic equation: $n^2 + 2n - 48 = 0$ Factor the quadratic equation: We need two numbers that multiply to -48 and add to 2. These numbers are 8 and -6. $(n+8)(n-6) = 0$ This gives two possible values for $n$: $n = -8 \quad \text{or} \quad n = 6$

5. Validate the Solution

In the context of the problem, $n$ represents the upper limit of a summation $\sum_{k=1}^n$. This implies that $n$ must be a positive integer.

• $n = -8$ is not a positive integer, so we reject this solution.
• $n = 6$ is a positive integer, so this is the valid solution.

Therefore, $n = 6$.

---

Summary and Key Takeaway

This problem effectively tests your understanding of determinant properties, specifically how summation interacts with determinants. The key steps were:

1. Recognizing and applying the summation property of determinants: This allows converting a sum of determinants into a single determinant.
2. Strategic use of row/column operations: Simplifying the determinant by creating zeros in a row or column is crucial for efficient calculation.
3. Careful algebraic manipulation: Solving the resulting quadratic equation accurately.
4. Contextual validation of the answer: Always remember to check if your mathematical solution makes sense in the problem's context (e.g., $n$ being a positive integer for a summation limit).

By following these steps, even complex-looking determinant problems can be systematically solved.