# Chapter 10 - Section 10.4 - Mathematical Induction - Exercise Set - Page 1085: 28

See the explanation below.

#### Work Step by Step

Let us consider the statement ${{S}_{n}}:3\text{ is a factor of }n\left( n+1 \right)\left( n-1 \right)$. 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-1 \right)=1\left( 1+1 \right)\left( 1-1 \right) \\ & =0 \end{align} Therefore, the above statement is true for $n=1$ as 3 is a factor of 0. So, the statement ${{S}_{1}}$ is true. Then, show that, if ${{S}_{k}}$ is true, then ${{S}_{k+1}}$ is true. And 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. ${{S}_{k}}:3\text{ is a factor of }k\left( k+1 \right)\left( k-1 \right)$ And, ${{S}_{k+1}}:3\text{ is a factor of }\left( k+1 \right)\left( k+1+1 \right)\left( k+1-1 \right)$ That is, ${{S}_{k+1}}:3\text{ is a factor of }k\left( k+1 \right)\left( k+2 \right)$ Here, we assume that ${{S}_{k}}$ is true, so add the next consecutive integer after $k$, specifically $3{{k}^{2}}+3k$ to both sides of the equation. $3\text{ is a factor of }k\left( k+1 \right)\left( k-1 \right)={{k}^{3}}-k$ And add $3{{k}^{2}}+3k$ on both sides, \begin{align} & 3\text{ is a factor of }k\left( k+1 \right)\left( k-1 \right)+3{{k}^{2}}+3k={{k}^{3}}-k+3{{k}^{2}}+3k \\ & 3\text{ is a factor of }k\left( k+1 \right)\left( k+2 \right)={{k}^{3}}+3{{k}^{2}}+2k \end{align} 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 $3{{k}^{2}}+3k$ to both sides of ${{S}_{k}}$, then ${{S}_{k+1}}$ is also true. Thus, by the principal of mathematical induction, the statement ${{S}_{n}}:3\text{ is a factor of }n\left( n+1 \right)\left( n-1 \right)$ is true for every positive integer $n$.

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