## Precalculus (6th Edition) Blitzer

\begin{align} & {{S}_{1}}:1\cdot 3=\frac{1\left( 1+1 \right)\left[ 2\left( 1 \right)+7 \right]}{6} \\ & 3=\frac{2\cdot 9}{6} \\ & 3=3\,\ \text{is true}\text{.} \\ \end{align} \begin{align} & {{S}_{k}}:1\cdot 3+2\cdot 4+3\cdot 5+....+k\left( k+2 \right)=\frac{k\left( k+1 \right)\left( 2k+7 \right)}{6} \\ & {{S}_{k+1}}:1\cdot 3+2\cdot 4+3\cdot 5+....+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} Then simplify the right-hand side: \begin{align} & 1\cdot 3+2\cdot 4+3\cdot 5+....+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) \\ & =\frac{k\left( k+1 \right)\left( 2k+7 \right)+6\left( k+1 \right)\left( k+3 \right)}{6} \\ & =\frac{\left( k+1 \right)\left[ k\left( 2k+7 \right)+6\left( k+3 \right) \right]}{6} \\ & =\frac{\left( k+1 \right)\left( 2{{k}^{2}}+13k+18 \right)}{6} \\ & =\frac{\left( k+1 \right)\left( k+2 \right)\left( 2k+9 \right)}{6} \end{align} If ${{S}_{k}}$ is true, then ${{S}_{k+1}}$ is true.