Answer
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 $.
