JEE Main 2026 Apr 5 Shift 2, Mathematics Q3: Differentiation And Intergration In Determinants

Since only the first column of the determinant depends on $n$, summing $f(n)$ from $n=1$ to $k$ replaces the first column entries $n,-2n^2,-3n^3$ by their sums:

$\sum n=\frac{k(k+1)}{2}$, $\sum -2n^2=-\frac{k(k+1)(2k+1)}{3}$, $\sum -3n^3=-\frac{3k^2(k+1{)}^2}{4}$.

Expanding the resulting determinant and simplifying yields:

$\sum _{n=1}^{k}f(n)=\frac{k(k+1)(2k+1)}{2}\cdot \frac{7}{3}$.

Setting this equal to $98$: $k(k+1)(2k+1)=84$. Testing $k=3$: $3\cdot 4\cdot 7=84$. $✓$

$\therefore k=3$.