Let $$D_k = \begin{vmatrix} 1 & 2k & 2k-1 \\ n & n^2+n+2 & n^2 \\ n & n^2+n & n^2+n+2 \end{vmatrix}$$. If $$\sum_{k=1}^{n} D_k = 96$$, then $$n$$ is equal to _____.

Explanation

Given,

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

Using

$$\sum_{k=1}^{n}1=n$$

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

$$\sum_{k=1}^{n}(2k-1)=n^2$$

we get

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

Applying the row operations $$R_2\rightarrow R_2-R_1$$ and $$R_3\rightarrow R_3-R_1$$,

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

Therefore,

$$\sum_{k=1}^{n} D_k=n\cdot2\cdot(n+2)=2n(n+2)$$

Given that

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

we have

$$2n(n+2)=96$$

$$n(n+2)=48$$

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

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

Hence,

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

Since $$n$$ is positive,

$$\boxed{n=6}$$