## Precalculus (6th Edition) Blitzer

We use mathematical induction as follows: Statement ${{S}_{1}}$ is ${{S}_{1}}:2$ And simplifying on the right, we obtain, \begin{align} & \frac{n\left( 5n-1 \right)}{2}=\frac{1\left( 5-1 \right)}{2} \\ & =\frac{4}{2} \\ & =2 \end{align}. So, this statement shows that ${{S}_{1}}$ is true. And assume ${{S}_{k}}$ is true. Using ${{S}_{k}},{{S}_{k+1}}$ from the expression, ${{S}_{k}}:2+7+12+...+\left( 5k-3 \right)=\frac{k\left( 5k-1 \right)}{2}$ Add $\left( 5\left( k+1 \right)-3 \right)$ on both sides, \begin{align} & 2+7+12+...+\left( 5k-3 \right)+\left( 5\left( k+1 \right)-3 \right)=\frac{k\left( 5k-1 \right)}{2}+\left( 5\left( k+1 \right)-3 \right) \\ & =\frac{k\left( 5k-1 \right)}{2}+\left( 5k+2 \right) \\ & =\frac{5{{k}^{2}}+9k+4}{2} \\ & =\frac{\left( 5{{k}^{2}}+4k+5k+4 \right)}{2} \end{align} We simplify further, \begin{align} & 2+7+12+...+\left( 5k-3 \right)+\left( 5\left( k+1 \right)-3 \right)=\frac{k\left( 5k+4 \right)+1\left( 5k+4 \right)}{2} \\ & =\frac{\left( k+1 \right)\left( 5k+4 \right)}{2} \\ & =\frac{\left( k+1 \right)\left( 5\left( k+1 \right)-1 \right)}{2} \end{align} Therefore, ${{S}_{k+1}}$ is true. The result ${{S}_{n}}:2+7+12+...+\left( 5n-3 \right)=\frac{n\left( 5n-1 \right)}{2}$ is true by mathematical induction. Thus, the provided statement is proved by mathematical induction.