## Precalculus (6th Edition) Blitzer

The required value of ${{S}_{1}}=3$, ${{S}_{2}}=7$ and ${{S}_{3}}=12$, and also the statement is true.
Let us c consider the statement: ${{S}_{n}}:3+4+5+\cdots +\left( n+2 \right)=\frac{n\left( n+5 \right)}{2}$ For ${{S}_{1}}$ one has \begin{align} & {{S}_{1}}:3=\frac{1\cdot \left( 1+5 \right)}{2} \\ & =3 \end{align} Therefore, the above statement is true for $n=1$. For ${{S}_{2}}$ one has \begin{align} & {{S}_{2}}:3+4=\frac{2\cdot \left( 2+5 \right)}{2} \\ & =7 \end{align} So, the above statement is true for $n=2$. For ${{S}_{3}}$ one has \begin{align} & {{S}_{3}}:3+4+5=\frac{3\cdot \left( 3+5 \right)}{2} \\ & =12 \end{align} Therefore, the above statement is true for $n=3$. Thus, the values we obtain are ${{S}_{1}}=3,\,\,{{S}_{2}}=7$, and ${{S}_{3}}=12$. And the statement is true.