Answer
See the explanation below.
Work Step by Step
$\begin{align}
& {{S}_{1}}:1\cdot 3=\frac{1\left( 1+1 \right)\left[ 2\left( 1 \right)+7 \right]}{6} \\
& 3=\frac{2\cdot 9}{6} \\
& 3=3\,\ \text{is true}\text{.} \\
\end{align}$
$\begin{align}
& {{S}_{k}}:1\cdot 3+2\cdot 4+3\cdot 5+....+k\left( k+2 \right)=\frac{k\left( k+1 \right)\left( 2k+7 \right)}{6} \\
& {{S}_{k+1}}:1\cdot 3+2\cdot 4+3\cdot 5+....+k\left( k+2 \right)+\left( k+1 \right)\left( k+3 \right)=\frac{\left( k+1 \right)\left( k+2 \right)\left( 2k+9 \right)}{6}
\end{align}$
Then simplify the right-hand side:
$\begin{align}
& 1\cdot 3+2\cdot 4+3\cdot 5+....+k\left( k+2 \right)+\left( k+1 \right)\left( k+3 \right)=\frac{k\left( k+1 \right)\left( 2k+7 \right)}{6}+\left( k+1 \right)\left( k+3 \right) \\
& =\frac{k\left( k+1 \right)\left( 2k+7 \right)+6\left( k+1 \right)\left( k+3 \right)}{6} \\
& =\frac{\left( k+1 \right)\left[ k\left( 2k+7 \right)+6\left( k+3 \right) \right]}{6} \\
& =\frac{\left( k+1 \right)\left( 2{{k}^{2}}+13k+18 \right)}{6} \\
& =\frac{\left( k+1 \right)\left( k+2 \right)\left( 2k+9 \right)}{6}
\end{align}$
If ${{S}_{k}}$ is true, then ${{S}_{k+1}}$ is true.
