JEE Main 2026 · 5 Apr · Shift 2, Mathematics question 3

Sum column by column. Only the first column depends on n, so $\sum f(n)$ is the same determinant with the first column replaced by $\left(S_1,\,-2S_2,\,-3S_3\right)$, where
$$S_1=\tfrac{k(k+1)}2,\qquad S_2=\tfrac{S_1(2k+1)}3,\qquad S_3=S_1^2.$$

Cofactors of the first column.

• $C_{11}=3(2k+1)(k^2+5k+1)$.
• $C_{21}=-(27k^2+9k-1)$.
• $C_{31}=14(2k+1)$.

Expand.
$$\sum f=S_1(2k+1)\left[3(k^2+5k+1)+\tfrac23(27k^2+9k-1)-42S_1\right]=\frac73S_1(2k+1)=\frac{7k(k+1)(2k+1)}6.$$

Solve. $\dfrac{7k(k+1)(2k+1)}6=98$ gives $k(k+1)(2k+1)=84$, so $k=3$.

(A)