JEE Main 20265 Apr, Evening shiftMathematicsMediumMatrices and Determinants245× since 2002If f:N→Zf:\mathbb{N}\to\mathbb{Z}f:N→Z is defined by f(n)=∣n−1−5−2n23(2k+1)2k+1−3n33k(2k+1)3k(k+2)+1∣f(n)=\begin{vmatrix} n & -1 & -5 \\ -2n^2 & 3(2k+1) & 2k+1 \\ -3n^3 & 3k(2k+1) & 3k(k+2)+1 \end{vmatrix}f(n)=n−2n2−3n3−13(2k+1)3k(2k+1)−52k+13k(k+2)+1, k∈Nk\in\mathbb{N}k∈N, and ∑n=1kf(n)=98\sum_{n=1}^{k} f(n)=98∑n=1kf(n)=98, then kkk is equal to :AOption A. 3BOption B. 4COption C. 5DOption D. 6CheckSkip · show solution
