If $\mathrm{f}: \mathrm{N} \rightarrow \mathrm{Z}$ is defined by $f(n)=\left|\begin{array}{ccc}n & -1 & -5 \\ -2 n^{2} & 3(2 k+1) & 2 k+1 \\ -3 n^{3} & 3 k(2 k+1) & 3 k(k+2)+1\end{array}\right|, k \in N$, and $\sum_{n=1}^{k} f(n)=98$, then $k$ is equal to:

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Determinants

f : N \to Z

is defined by

f (n) = n - 2 n^{2} - 3 n^{3} - 1 3 (2 k + 1) 3 k (2 k + 1) - 5 2 k + 1 3 k (k + 2) + 1, k \in N

, and

\sum_{n = 1}^{k} f (n) = 98

, then

k

is equal to:

Answer: (A)

The rows are linear in the first-column entries, so the sum over

n

may be taken column by column, replacing

(n, - 2 n^{2}, - 3 n^{3})^{T}

(\sum n, - 2 \sum n^{2}, - 3 \sum n^{3})^{T}

while the other two columns stay fixed. Evaluating the determinant reduces it to

n = 1 \sum k f (n) = 7 n = 1 \sum k n^{2} = \frac{7 k ( k + 1 ) ( 2 k + 1 )}{6}

. Setting this equal to

98

gives

k (k + 1) (2 k + 1) = 84

, and

3 \cdot 4 \cdot 7 = 84

. Hence

k = 3

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Single correct option