Precalculus (6th Edition) Blitzer

We use mathematical induction as follows: Statement ${{S}_{1}}$ is: ${{S}_{1}}:3$ Then, simplify on the right and obtain, \begin{align} & n\left( 2n+1 \right)=1\left( 2+1 \right) \\ & =3 \end{align}. This statement shows that ${{S}_{1}}$ is true. Suppose ${{S}_{k}}$ is true. By using ${{S}_{k}},{{S}_{k+1}}$ from the expression, ${{S}_{k}}:3+7+11+...+\left( 4k-1 \right)=k\left( 2k+1 \right)$ Add $\left( 4\left( k+1 \right)-1 \right)$ on both sides, \begin{align} & 3+7+11+...+\left( 4k-1 \right)+\left( 4\left( k+1 \right)-1 \right)=k\left( 2k+1 \right)+\left( 4\left( k+1 \right)-1 \right) \\ & =2{{k}^{2}}+5k+3 \\ & =2{{k}^{2}}+3k+2k+3 \\ & =k\left( 2k+3 \right)+1\left( 2k+3 \right) \end{align} Simplify \begin{align} & 3+7+11+...+\left( 4k-1 \right)+\left( 4\left( k+1 \right)-1 \right)=\left( k+1 \right)\left( 2k+3 \right) \\ & =\left( k+1 \right)\left( 2\left( k+1 \right)+1 \right) \end{align} Therefore, ${{S}_{k+1}}$ is true. The result ${{S}_{n}}:3+7+11+...+\left( 4n-1 \right)=n\left( 2n+1 \right)$ is true by mathematical induction. Thus, the provided statement is proved by mathematical induction.