#### Answer

See the explanation below.

#### Work Step by Step

We use mathematical induction as follows:
Statement ${{S}_{1}}$ is:
$\begin{align}
& {{n}^{2}}-n={{1}^{2}}-1 \\
& =0
\end{align}$
$2$ is a factor of $0$
So, this statement shows that ${{S}_{1}}$ is true.
Also, assume ${{S}_{k}}$ is true. Using ${{S}_{k}},{{S}_{k+1}}$ from the expression,
${{S}_{k}}:2$ is a factor of ${{k}^{2}}-k $
To show, ${{S}_{k+1}}:2$ is a factor of ${{\left( k+1 \right)}^{2}}-\left( k+1 \right)$
Consider,
$\begin{align}
& {{\left( k+1 \right)}^{2}}-\left( k+1 \right)={{k}^{2}}+2k+1-k-1 \\
& ={{k}^{2}}+k \\
& =\left( {{k}^{2}}-k \right)+2k
\end{align}$
Since, 2 is a factor of ${{k}^{2}}-k $ and 2k, hence it is a factor of $\left( {{k}^{2}}-k \right)+2k $
So, 2 is a factor of ${{\left( k+1 \right)}^{2}}-\left( k+1 \right)$.
Therefore, ${{S}_{k+1}}$ is true. The result ${{S}_{n}}:2$ is a factor of ${{n}^{2}}-n $ is true by mathematical induction.
Thus, the provided statement is proved by mathematical induction.