JEE Main 2026 — Determinants Question with Solution

Let $S=\sum _{n=1}^{k}f(n)$, where
$f(n)=\left|\begin{array}{ccc}n& -1& -5\\ -2n^2& 3(2k+1)& 2k+1\\ -3n^3& 3k(2k+1)& 3k(k+2)+1\end{array}\right|$

Since only the first column depends on $n$, the summation can be taken inside that column:
$S=\left|\begin{array}{ccc}\sum _{n=1}^{k}n& -1& -5\\ \sum _{n=1}^k(-2n^2)& 3(2k+1)& 2k+1\\ \sum _{n=1}^k(-3n^3)& 3k(2k+1)& 3k(k+2)+1\end{array}\right|$

Using standard summation formulas:
$\sum _{n=1}^{k}n=\frac{k(k+1)}{2}$
$\sum _{n=1}^k(-2n^2)=-\frac{k(k+1)(2k+1)}{3}$
$\sum _{n=1}^k(-3n^3)=-\frac{3k^2(k+1{)}^2}{4}$

Substituting:
$S=\left|\begin{array}{ccc}\frac{k(k+1)}{2}& -1& -5\\ -\frac{k(k+1)(2k+1)}{3}& 3(2k+1)& 2k+1\\ -\frac{3k^2(k+1{)}^2}{4}& 3k(2k+1)& 3k^2+6k+1\end{array}\right|$

Taking $\frac{k(k+1)}{12}$ common from $C_1$:
$S=\frac{k(k+1)}{12}\left|\begin{array}{ccc}6& -1& -5\\ -4(2k+1)& 3(2k+1)& 2k+1\\ -9k(k+1)& 3k(2k+1)& 3k^2+6k+1\end{array}\right|$

Applying $C_1\to C_1+C_2+C_3$:
Row 1: $6-1-5=0$
Row 2: $-8k-4+6k+3+2k+1=0$
Row 3: $-9k^2-9k+6k^2+3k+3k^2+6k+1=1$

$S=\frac{k(k+1)}{12}\left|\begin{array}{ccc}0& -1& -5\\ 0& 3(2k+1)& 2k+1\\ 1& 3k(2k+1)& 3k^2+6k+1\end{array}\right|$

Expanding along $C_1$:
$S=\frac{k(k+1)}{12}\cdot 1\cdot \left|\begin{array}{cc}-1& -5\\ 3(2k+1)& 2k+1\end{array}\right|$

$S=\frac{k(k+1)}{12}\left[(-1)(2k+1)-(-5)(3(2k+1))\right]$

$S=\frac{k(k+1)}{12}\left[-(2k+1)+15(2k+1)\right]$

$S=\frac{k(k+1)}{12}\cdot 14(2k+1)=7\cdot \frac{k(k+1)(2k+1)}{6}$

Note that $\frac{k(k+1)(2k+1)}{6}=\sum _{n=1}^k{n}^2$.

Given $S=98$:
$7\cdot \frac{k(k+1)(2k+1)}{6}=98$

$\frac{k(k+1)(2k+1)}{6}=14$

Checking integer values of $k$:
$k=1$: sum $=1$
$k=2$: sum $=5$
$k=3$: sum $=14$

Hence, $k=3$, and the correct option is $(1)3$.