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 :

A

$$3$$

B

$$4$$

C

$$5$$

D

$$6$$