Let $$f : \mathbb{N} \to \mathbb{Z}$$ be defined by $$f(n) = \det\begin{bmatrix} n  & -1 & -5\\-2n^2 & 3(2k+1) & 2k+1 \\ -3n^3 & 3k(2k+1) & 3k(k+2)+1 \end{bmatrix}$$, $$k \in \mathbb{N}$$ and  $$\displaystyle\sum_{n=1}^{k} f(n) = 98$$, then $$k$$ is equal to :

$$\sum_{n=1}^k f(n) = \text{det} \begin{bmatrix} \sum_{n=1}^k n & -1 & -5 \\ \sum_{n=1}^k -2n^2 & 3(2k+1) & 2k+1 \\ \sum_{n=1}^k -3n^3 & 3k(2k+1) & 3k(k+2)+1 \end{bmatrix}$$

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

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

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

$$\sum_{n=1}^k f(n) = \frac{k(k+1)}{12} \text{det} \begin{bmatrix} 6 & -1 & -5 \\ -4(2k+1) & 3(2k+1) & 2k+1 \\ -9k(k+1) & 3k(2k+1) & 3k^2+6k+1 \end{bmatrix}$$

$$C_1 \rightarrow C_1 + 6C_2$$:

$$\sum_{n=1}^k f(n) = \frac{k(k+1)}{12} \text{det} \begin{bmatrix} 0 & -1 & -5 \\ 14(2k+1) & 3(2k+1) & 2k+1 \\ 9k(2k+1) & 3k(2k+1) & 3k^2+6k+1 \end{bmatrix}$$

$$C_3 \rightarrow C_3 - 5C_2$$:

$$\sum_{n=1}^k f(n) = \frac{k(k+1)}{12} \text{det} \begin{bmatrix} 0 & -1 & 0 \\ 14(2k+1) & 3(2k+1) & -14(2k+1) \\ 9k(2k+1) & 3k(2k+1) & 3k^2-24k-14 \end{bmatrix}$$

$$\sum_{n=1}^k f(n) = \frac{k(k+1)}{12} \cdot (-1)(-1)^{1+2} \cdot \text{det} \begin{bmatrix} 14(2k+1) & -14(2k+1) \\ 9k(2k+1) & 3k^2-24k-14 \end{bmatrix}$$

$$\sum_{n=1}^k f(n) = -\frac{k(k+1)(2k+1)}{12} \cdot 14 \cdot \left[ (3k^2-24k-14) - (-9k) \right]$$

$$\sum_{n=1}^k f(n) = -\frac{14k(k+1)(2k+1)(3k^2-15k-14)}{12}$$

If $$k = 3$$: $$\sum_{n=1}^3 f(n) = f(1) + f(2) + f(3) = 98$$

Hence, option (A) is correct.