## Precalculus (6th Edition) Blitzer

We use mathematical induction as follows: Statement ${{S}_{1}}$ is; $1\cdot 2=2$ And simplifying on the right, we obtain $\frac{n\left( n+1 \right)\left( n+2 \right)}{3}=2$. 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, $1\cdot 2+2\cdot 3+...+k\left( k+1 \right)=\frac{k\left( k+1 \right)\left( k+2 \right)}{3}$ By adding $\left( k+1 \right)\left( k+2 \right)$ on both sides, we get: \begin{align} & 1\cdot 2+2\cdot 3+...+k\left( k+1 \right)+\left( k+1 \right)+\left( k+2 \right)=\frac{k\left( k+1 \right)\left( k+2 \right)}{3}+\left( k+1 \right)\left( k+2 \right) \\ & 1\cdot 2+2\cdot 3+...+k\left( k+1 \right)+\left( k+1 \right)+\left( k+2 \right)=\frac{k\left( k+1 \right)\left( k+2 \right)+3\left( k+1 \right)\left( k+2 \right)}{3} \\ & 1\cdot 2+2\cdot 3+...+k\left( k+1 \right)+\left( k+1 \right)+\left( k+2 \right)=\frac{\left( k+1 \right)\left( k+2 \right)\left( k+3 \right)}{3} \\ \end{align} Thus, ${{S}_{k+1}}$ is true. The result $1\cdot 2+2\cdot 3+...+n\left( n+1 \right)=\frac{n\left( n+1 \right)\left( n+2 \right)}{3}$ holds true.