## Precalculus (6th Edition) Blitzer

We use mathematical induction as follows: Statement ${{S}_{1}}$ is; $1\cdot 3=3$ And simplifying on the right, we obtain $\frac{n\left( n+1 \right)\left( 2n+7 \right)}{6}=3$. So, this true statement shows that ${{S}_{1}}$ is true. Also, assume ${{S}_{k}}$ is true. Using ${{S}_{k}},{{S}_{k+1}}$ from the expression, $1\cdot 3+2\cdot 4+...+k\left( k+2 \right)=\frac{k\left( k+1 \right)\left( 2k+7 \right)}{6}$ By adding $\left( k+1 \right)\left( k+3 \right)$ on both sides, we get: \begin{align} & 1\cdot 3+2\cdot 4+...+k\left( k+2 \right)+\left( k+1 \right)\left( k+3 \right)=\frac{k\left( k+1 \right)\left( 2k+7 \right)}{6}+\left( k+1 \right)\left( k+3 \right) \\ & 1\cdot 3+2\cdot 4+...+k\left( k+2 \right)+\left( k+1 \right)\left( k+3 \right)=\frac{k\left( k+1 \right)\left( 2k+7 \right)+6\left( k+1 \right)\left( k+3 \right)}{6} \\ & 1\cdot 3+2\cdot 4+...+k\left( k+2 \right)+\left( k+1 \right)\left( k+3 \right)=\frac{k\left( k+1 \right)\left( 2k+7 \right)+6\left( k+1 \right)\left( k+3 \right)}{6} \\ & 1\cdot 3+2\cdot 4+...+k\left( k+2 \right)+\left( k+1 \right)\left( k+3 \right)=\frac{2{{k}^{3}}+9{{k}^{2}}+7k+6{{k}^{2}}+24k+18}{6} \\ \end{align} Therefore, \begin{align} & 1\cdot 3+2\cdot 4+...+k\left( k+2 \right)+\left( k+1 \right)\left( k+3 \right)=\frac{2{{k}^{3}}+15{{k}^{2}}+31k+18}{6} \\ & 1\cdot 3+2\cdot 4+...+k\left( k+2 \right)+\left( k+1 \right)\left( k+3 \right)=\frac{\left( k+1 \right)\left( k+2 \right)\left( 2k+9 \right)}{6} \\ \end{align} Thus, ${{S}_{k+1}}$ is true. The result $1\cdot 3+2\cdot 4+...+n\left( n+2 \right)=\frac{n\left( n+1 \right)\left( 2n+7 \right)}{6}$ holds true.