## Precalculus (6th Edition) Blitzer

Now, consider the statement: $2$ is a factor of ${{n}^{2}}+3n$ If $n=1$, the statement is $2$ is a factor of $1+3$ If $n=2$, the statement is $2$ is a factor of $4+6$ If $n=3$, the statement is $2$ is a factor of $9+9$ If $n=k+1$, the statement before the algebra is simplified is $2$ is a factor of ${{\left( k+1 \right)}^{2}}+3\left( k+1 \right)$ If $n=k+1$, the provided statement after the algebra is simplified is $2$ is a factor of ${{k}^{2}}+5k+4$
For the statement $2$ is a factor of ${{n}^{2}}+3n$: if $n=1$ substituting, \begin{align} & {{n}^{2}}+3n={{1}^{2}}+3\times 1 \\ & =1+3 \\ & =4 \end{align} If the value of $n=2$: \begin{align} & {{n}^{2}}+3n={{2}^{2}}+3\times 2 \\ & =4+6 \\ & =10 \end{align} Then, the value $n=3$ gives \begin{align} & {{n}^{2}}+3n={{3}^{2}}+3\times 3 \\ & =9+9 \\ & =18 \end{align} If $n=k+1$, $2$ is a factor ${{\left( k+1 \right)}^{2}}+3\left( k+1 \right)$ Further simplification gives, \begin{align} & {{\left( k+1 \right)}^{2}}+3\left( k+1 \right)={{k}^{2}}+2k+1+3k+3 \\ & ={{k}^{2}}+5k+4 \end{align}