Explanation

(a)
$\sum_{n = 1}^{N}U_{n} = \left| \begin{matrix}
\sum_{}^{}n & 1 & 5 \\
\sum_{}^{}n^{2} & 2N + 1 & 2N + 1 \\
\sum_{}^{}n^{3} & 3N^{2} & 3N
\end{matrix} \right|$
$= \left| \begin{matrix}
\frac{N(N + 1)}{2} & 1 & 5 \\
\frac{N(N + 1)(2N + 1)}{6} & 2N + 1 & 2N + 1 \\
\left( \ \frac{N(N + 1)}{2} \right)^{2} & 3N^{2} & 3N
\end{matrix} \right|$
$= \frac{N(N + 1)}{12}\left| \begin{matrix}
6 & 1 & 5 \\
4N + 2 & 2N + 1 & 2N + 1 \\
3N(N + 1) & 3N^{2} & 3N
\end{matrix} \right|$
Applying $C_{3} \rightarrow C_{3} + C_{2}$
$= \frac{N(N + 1)}{12}\left| \begin{matrix}
6 & 1 & 6 \\
4N + 2 & 2N + 1 & 4N + 2 \\
3N(N + 1) & 3N^{2} & 3N(N + 1)
\end{matrix} \right|$
$= 0\ \ (\because\$two columns are identical)