## Precalculus (6th Edition) Blitzer

We use mathematical induction as follows: Statement ${{S}_{1}}$ is; $3n=3$ Further simplifying on the right, we obtain $\frac{3n\left( n+1 \right)}{2}=3$. This true statement shows that ${{S}_{1}}$ is true. Suppose ${{S}_{k}}$ is true. Using ${{S}_{k}},{{S}_{k+1}}$ from the expression, ${{S}_{k}}=3+6+9+....+3k=\frac{3k\left( k+1 \right)}{2}$ Adding $3\left( k+1 \right)$ on both sides as given below: \begin{align} & 3+6+9+\ldots +3k+3\left( k+1 \right)=\frac{3k\left( k+1 \right)}{2}+3\left( k+1 \right) \\ & 3+6+9+\ldots +3k+3\left( k+1 \right)=\frac{3k\left( k+1 \right)+6\left( k+1 \right)}{2} \\ & 3+6+9+\ldots +3k+3\left( k+1 \right)=\frac{3{{k}^{2}}+3k+6k+6}{2} \\ & 3+6+9+\ldots +3k+3\left( k+1 \right)=\frac{3{{k}^{2}}+9k+6}{2} \end{align} Therefore, $3+6+9+\ldots +3k+3\left( k+1 \right)=\frac{3\left( k+1 \right)\left( k+2 \right)}{2}$ Thus, ${{S}_{k+1}}$ is true. The result ${{S}_{n}}=3+6+9+....+3n=\frac{3n\left( n+1 \right)}{2}$ holds true.