If $f: \mathbf{N} \rightarrow \mathbf{Z}$ is defined by
$$
f(n)=\left|\begin{array}{ccc}
n & -1 & -5 \\
-2 n^2 & 3(2 k+1) & 2 k+1 \\
-3 n^3 & 3 k(2 k+1) & 3 k(k+2)+1
\end{array}\right|, k \in N,
$$
and $\sum_{n=1}^k f(n)=98$, then $k$ is equal to:
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