Explanation

(c)
Putting $r = 1,\ 2,\ 3,\ldots,n$ and using the formula
$\sum_{}^{}{1 = n\ and\ \sum_{}^{}{r = \frac{(n + 1)n}{2}}}$
$\sum_{}^{}{(2r - 1) = 1 + 3 + 5 + ... = n^{2}}$
$\therefore\ \sum_{r = 1}^{n}{\mathrm{\Delta}_{r} =}\left| \begin{matrix}
n & n & n \\
n(n + 1) & n^{2} + n + 1 & n^{2} + n \\
n^{2} & n^{2} & n^{2} + n + 1
\end{matrix} \right| = 56$
Applying $C_{1} \rightarrow C_{1} - C_{3},\ C_{2} \rightarrow C_{2} - C_{3}$
$\left| \begin{matrix}
0 & 0 & n \\
0 & 1 & n^{2} + n \\
- n - 1 & - n - 1 & n^{2} + n + 1
\end{matrix} \right|$
$\Rightarrow n(n + 1) = 56$
$\Rightarrow n^{2} + n - 56 = 0$
$\Rightarrow (n + 8)(n - 7) = 0$
$\Rightarrow n = 7\ \ \ \ \ (n \neq - 8)$