## Precalculus (6th Edition) Blitzer

We use mathematical induction as follows: Statement ${{S}_{1}}$ is ${{S}_{1}}:4$ We are solving the right side as, \begin{align} & 2n\left( n+1 \right)=2\left( 1 \right)\left( 1+1 \right) \\ & =2\left( 2 \right) \\ & =4 \end{align} This statement shows that ${{S}_{1}}$ is true. Suppose ${{S}_{k}}$ is true. Using ${{S}_{k}},{{S}_{k+1}}$ from the expression, \begin{align} & {{S}_{k}}:4+8+12+...+4k=2k\left( k+1 \right) \\ & \text{ }=2{{k}^{2}}+2 \\ \end{align} Then we have ${{S}_{k+1}}$ ${{S}_{k+1}}:$ \begin{align} & 4+8+12+...+4\left( k+1 \right)=2\left( k+1 \right)\left( k+1+1 \right) \\ & =2\left( k+1 \right)\left( k+2 \right) \\ & =2\left( {{k}^{2}}+k+2k+2 \right) \\ & =\left( 2{{k}^{2}}+6k+4 \right) \end{align} Solving further \begin{align} & 4+8+12+...+4\left( k+1 \right)=2\left( {{k}^{2}}+3k+1 \right)+2 \\ & =2{{\left( k+1 \right)}^{2}}+2 \end{align} Thus, ${{S}_{k+1}}$ is true. The result ${{S}_{n}}:4+8+12+...+4n=2n\left( n+1 \right)$ is true by mathematical induction. Hence, the provided statement is proved by mathematical induction.