JEE Main 2026 — 5 April, Evening Shift — Question 27

$f(n)=\left|\begin{array}{ccc}n & -1 & -5\\ -2 n^{2} & 3(2 k+1) & 2 k+1\\ -3 n^{3} & 3 k(2 k+1) & 3 k(k+2)+1\end{array}\right|$ $\sum_{n=1}^{k} f(n)=\left|\begin{array}{ccc}\frac{k(k+1)}{2} & -1 & -5\\ -2 \frac{k(k+1)(2 k+1)}{6} & 3(2 k+1) & 2 k+1\\ -3 \frac{k^{2}(k+1)^{2}}{4} & 3 k(2 k+1) & 3 k(k+2)+1\end{array}\right|=98$ $\Rightarrow \frac{\mathrm{k}(\mathrm{k}+1)(2 \mathrm{k}+1)}{2} \cdot \frac{7}{3}=98$ $\Rightarrow \mathrm{k}(\mathrm{k}+1)(2 \mathrm{k}+1)=84$ $\Rightarrow \mathrm{k}=3$