If f: N arrow Z is defined by f(n) = vmatrix n & -1 & -5 -2n^2 &...

Let $S = \displaystyle\sum_{n=1}^{k} f(n)$, where
$f(n) = \begin{vmatrix} n & -1 & -5 \\ -2n^2 & 3(2k+1) & 2k+1 \\ -3n^3 & 3k(2k+1) & 3k(k+2)+1 \end{vmatrix}$

Since only the first column depends on $n$, the summation can be taken inside that column:
$S = \begin{vmatrix} \displaystyle\sum_{n=1}^{k} n & -1 & -5 \\ \displaystyle\sum_{n=1}^{k} (-2n^2) & 3(2k+1) & 2k+1 \\ \displaystyle\sum_{n=1}^{k} (-3n^3) & 3k(2k+1) & 3k(k+2)+1 \end{vmatrix}$

Using standard summation formulas:
$\displaystyle\sum_{n=1}^{k} n = \dfrac{k(k+1)}{2}$
$\displaystyle\sum_{n=1}^{k} (-2n^2) = -\dfrac{k(k+1)(2k+1)}{3}$
$\displaystyle\sum_{n=1}^{k} (-3n^3) = -\dfrac{3k^2(k+1)^2}{4}$

Substituting:
$S = \begin{vmatrix} \dfrac{k(k+1)}{2} & -1 & -5 \\ -\dfrac{k(k+1)(2k+1)}{3} & 3(2k+1) & 2k+1 \\ -\dfrac{3k^2(k+1)^2}{4} & 3k(2k+1) & 3k^2+6k+1 \end{vmatrix}$

Taking $\dfrac{k(k+1)}{12}$ common from $C_1$:
$S = \dfrac{k(k+1)}{12} \begin{vmatrix} 6 & -1 & -5 \\ -4(2k+1) & 3(2k+1) & 2k+1 \\ -9k(k+1) & 3k(2k+1) & 3k^2+6k+1 \end{vmatrix}$

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: $-9k^2 - 9k + 6k^2 + 3k + 3k^2 + 6k + 1 = 1$

$S = \dfrac{k(k+1)}{12} \begin{vmatrix} 0 & -1 & -5 \\ 0 & 3(2k+1) & 2k+1 \\ 1 & 3k(2k+1) & 3k^2+6k+1 \end{vmatrix}$

Expanding along $C_1$:
$S = \dfrac{k(k+1)}{12} \cdot 1 \cdot \begin{vmatrix} -1 & -5 \\ 3(2k+1) & 2k+1 \end{vmatrix}$

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

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

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

Note that $\dfrac{k(k+1)(2k+1)}{6} = \displaystyle\sum_{n=1}^{k} n^2$.

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

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

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 $(1)\ 3$.