## Precalculus (6th Edition) Blitzer

The value of ${{S}_{1}}=1$, ${{S}_{2}}=4$ and ${{S}_{3}}=9$, and also the statement is true.
We consider the statement: ${{S}_{n}}:1+3+5+\cdots +\left( 2n-1 \right)={{n}^{2}}$ For ${{S}_{1}}$ we has ${{S}_{1}}:1={{1}^{2}}$ Therefore, the above statement is true for $n=1$. For ${{S}_{2}}$ one has ${{S}_{2}}:1+3={{2}^{2}}$ Therefore, the above statement is true for $n=2$. For ${{S}_{3}}$ one has ${{S}_{3}}:1+3+5={{3}^{2}}$ Therefore, the above statement is true for $n=3$. Thus, the statement holds. Thus the values are ${{S}_{1}}=1,\,\,{{S}_{2}}=4$, and ${{S}_{3}}=9$. And the statement is true.