## Precalculus (6th Edition) Blitzer

The required value of ${{S}_{1}}=0$, ${{S}_{2}}=2$ and ${{S}_{3}}=6$, and also the statement is true.
Let us consider the statement: ${{S}_{n}}:2\text{ is a factor of }{{n}^{2}}-n$. For ${{S}_{1}}$ one has \begin{align} & {{S}_{1}}:{{n}^{2}}-n={{1}^{2}}-1 \\ & =0 \end{align} Therefore, the above statement is true for $n=1$ as 2 is a factor of 0. For ${{S}_{2}}$ one has \begin{align} & {{S}_{2}}:{{2}^{2}}-2=4-2 \\ & =2 \end{align} Now, the above statement is true for $n=2$ as 2 is a factor o f2. For ${{S}_{3}}$ one has \begin{align} & {{S}_{3}}:{{3}^{2}}-3=9-3 \\ & =6 \end{align} Thus, the above statement is true for $n=3$ as 2 is a factor of 6. The values are ${{S}_{1}}=0,\,\,{{S}_{2}}=2$, and ${{S}_{3}}=6$. Also, the statement is true.