## Precalculus (6th Edition) Blitzer

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.