Answer
See the explanation below.
Work Step by Step
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.
