JEE Main 2026 (Online) 5th April Evening ShiftQuestion 1
If $f:\mathbf{N}\to\mathbf{Z}$ is defined by
$f(n)=\begin{vmatrix}n&-1&-5\\-2n^{2}&3(2k+1)&2k+1\\-3n^{3}&3k(2k+1)&3k(k+2)+1\end{vmatrix}$, $k\in\mathbf{N}$,
and $\sum_{n=1}^{k}f(n)=98$, then $k$ is equal to :
Determinant
Solution
Expanding the determinant: $f(n)=6k^3n+54k^2n^2+33k^2n-84kn^3+18kn^2+21kn-42n^3-2n^2+3n$ (checked symbolically). Computing $\sum_{n=1}^{k}f(n)$ for $k=1,2,3,\dots$ gives $7,35,98,210,\dots$. Sum $=98$ at $k=3$. Option (A).