If f: N → Z is defined by f(n)= |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|, k ∈ N, and Σ_n =1^k f( n )=98, then k is equal…

If $\displaystyle f: \mathbf{N} \rightarrow \mathbf{Z}$ 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 $\displaystyle \sum_{\mathrm{n}=1}^{\mathrm{k}} f(\mathrm{n})=98$, then k is equal to: