If f:N⟶Zf: N \longrightarrow Zf:N⟶Z is defined by f(n)=∣n−1−5−2n23(2k+1)2k+1−3n33k(2k+1)3k(k+2)+1∣,k∈Nf(n)=\begin{vmatrix} n & -1 & -5 \\ -2n^{2} & 3(2k+1) & 2k+1 \\ -3n^{3} & 3k(2k+1) & 3k(k+2)+1 \end{vmatrix}, k \in Nf(n)=n−2n2−3n3−13(2k+1)3k(2k+1)−52k+13k(k+2)+1,k∈N , and ∑n=1kf(n)=98\sum_{n=1}^{k} f(n)=98∑n=1kf(n)=98, then k is equal to:
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