Answer
See the explanation below.
Work Step by Step
We use mathematical induction as follows:
Statement ${{S}_{1}}$ is;
$\begin{align}
& {{n}^{2}}+3n={{1}^{2}}+3 \\
& =4
\end{align}$
$2$ is a factor of $4$
So, this true 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}}+3k $
And to show, ${{S}_{k+1}}:2$ is a factor of ${{\left( k+1 \right)}^{2}}+3\left( k+1 \right)$
$\begin{align}
& {{\left( k+1 \right)}^{2}}+3\left( k+1 \right)={{k}^{2}}+2k+1+3k+3 \\
& ={{k}^{2}}+3k+2\left( k+2 \right)
\end{align}$
Thus, ${{S}_{k+1}}$ is true. The result ${{S}_{n}}:2$ is a factor of ${{n}^{2}}+3n $ holds true.
