## Precalculus (6th Edition) Blitzer

The required solution is ${{S}_{k}}:2$ a factor of ${{k}^{2}}-k+2$, ${{S}_{k+1}}:2$ is a factor of ${{k}^{2}}+k+2$,
By using ${{S}_{k}},{{S}_{k+1}}$ from the expression ${{S}_{n}}:2$ is a factor of ${{n}^{2}}-n+2$ ${{S}_{k}}:$ 2 is a factor of ${{k}^{2}}-k+2$. ${{S}_{k+1}}:$ 2 is a factor of ${{\left( k+1 \right)}^{2}}-\left( k+1 \right)+2$ On solving further, \begin{align} & {{\left( k+1 \right)}^{2}}-\left( k+1 \right)+2={{k}^{2}}+2k+1-k-1+2 \\ & ={{k}^{2}}+k+2 \end{align} Thus, the solution is, ${{S}_{k}}:2$ is a factor of ${{k}^{2}}-k+2$, ${{S}_{k+1}}:2$ is a factor of ${{k}^{2}}+k+2$..