Answer
See the explanation below.
Work Step by Step
We use mathematical induction as follows:
Statement ${{S}_{1}}$ is;
$\frac{1}{2\cdot 3}=\frac{1}{6}$
And simplifying on the right, we obtain $\frac{n}{2n+4}=\frac{1}{6}$.
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,
$\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+...+\frac{1}{\left( k+1 \right)\left( k+2 \right)}=\frac{k}{2k+4}$
By adding $\frac{1}{\left( k+2 \right)\left( k+3 \right)}$ on both sides as:
$\begin{align}
& \frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+...+\frac{1}{\left( k+1 \right)\left( k+2 \right)}+\frac{1}{\left( k+2 \right)\left( k+3 \right)}=\frac{k}{2k+4}+\frac{1}{\left( k+2 \right)\left( k+3 \right)} \\
& \frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+...+\frac{1}{\left( k+1 \right)\left( k+2 \right)}+\frac{1}{\left( k+2 \right)\left( k+3 \right)}=\frac{k\left( k+3 \right)}{2\left( k+2 \right)\left( k+3 \right)}+\frac{2}{2\left( k+2 \right)\left( k+3 \right)} \\
& \frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+...+\frac{1}{\left( k+1 \right)\left( k+2 \right)}+\frac{1}{\left( k+2 \right)\left( k+3 \right)}=\frac{{{k}^{2}}+3k+2}{2\left( k+2 \right)\left( k+3 \right)} \\
\end{align}$
Therefore, $\begin{align}
& \frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+...+\frac{1}{\left( k+1 \right)\left( k+2 \right)}+\frac{1}{\left( k+2 \right)\left( k+3 \right)}=\frac{\left( k+1 \right)}{2\left( k+3 \right)} \\
& \frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+...+\frac{1}{\left( k+1 \right)\left( k+2 \right)}+\frac{1}{\left( k+2 \right)\left( k+3 \right)}=\frac{\left( k+1 \right)}{2\left( k+1 \right)+4} \\
\end{align}$
Thus, ${{S}_{k+1}}$ is true. The result $\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+...+\frac{1}{\left( n+1 \right)\left( n+2 \right)}=\frac{n}{2n+4}$ holds true.
