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

  • A
    $3$
  • B
    $4$
  • C
    $5$
  • D
    $6$

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