## Precalculus (6th Edition) Blitzer

Let us consider the statement ${{S}_{n}}:6\text{ is a factor of }n\left( n+1 \right)\left( n+2 \right)$. Firstly, show that ${{S}_{1}}$ is true. And write ${{S}_{1}}$ by taking the first term on the left and replacing n with $1$ on the right. \begin{align} & n\left( n+1 \right)\left( n+2 \right)=1\left( 1+1 \right)\left( 1+2 \right) \\ & =6 \end{align} Therefore, the above statement is true for $n=1$ as 6 is a factor of 6. So, the statement ${{S}_{1}}$ is true. Next, show that, if ${{S}_{k}}$ is true, then ${{S}_{k+1}}$ is true. Then write ${{S}_{k}}$ and ${{S}_{k+1}}$ by taking the sum of the first k and $\left( k+1 \right)$ terms on the left and replacing n with k and $\left( k+1 \right)$ on the right. \begin{align} & \text{ }k\left( k+1 \right)\left( k+2 \right)=k\left( {{k}^{2}}+3k+2 \right) \\ & ={{k}^{3}}+3{{k}^{2}}+2k \end{align} And, ${{S}_{k+1}}:6\text{ is a factor of }\left( k+1 \right)\left( k+1+1 \right)\left( k+1+2 \right)$ That is, ${{S}_{k+1}}:6\text{ is a factor of }\left( k+1 \right)\left( k+2 \right)\left( k+3 \right)={{k}^{3}}+6{{k}^{2}}+11k+6$ Here, we assume that ${{S}_{k}}$ is true, so add the next consecutive integer after $k$, specifically $2k$ to both sides of the equation. $6\text{ is a factor of }k\left( k+1 \right)\left( k+2 \right)$. And add $3{{k}^{2}}+9k+6$ on both sides, ${{S}_{k+1}}:6\text{ is a factor of }\left( k+1 \right)\left( k+2 \right)\left( k+3 \right)={{k}^{3}}+6{{k}^{2}}+11k+6$ So the final statement is ${{S}_{k+1}}$. Now, if we assume that ${{S}_{k}}$ is true, and add the next consecutive integer after k, specifically $2k$ to both sides of ${{S}_{k}}$, then ${{S}_{k+1}}$ is also true. Thus, by the principal of mathematical induction, the statement ${{S}_{n}}:6\text{ is a factor of }n\left( n+1 \right)\left( n+2 \right)$ is true for every positive integer $n$.