Answer
Prove that $p(1)$ is true.
Prove that $p(k)$ true implies $p(k+1)$ true.
Work Step by Step
Let's note by $p(n)$ the property we want to prove:
$$p(n):\text{ } \dfrac{1}{1\cdot 3}+\dfrac{1}{3\cdot 5}+\dfrac{1}{5\cdot 7}+\cdots+\dfrac{1}{(2n-1)(2n+1)}=\dfrac{n}{2n+1}.$$
$\textbf{Step 1:}$ Prove $p(1)$ is true.
Set $n=1$ on each side and check if the property is verified:
$$\begin{align*}
\dfrac{1}{(2(1)-1)(2(1)+1)}&\stackrel{?}{=}\dfrac{1}{(2(1+1)}\\
\dfrac{1}{1\cdot 3}&\stackrel{?}{=}\dfrac{1}{3}\\
\dfrac{1}{3}&=\dfrac{1}{3}\checkmark.
\end{align*}$$
Therefore $p(1)$ is true.
$\textbf{Step 2:}$ Prove that if $p(k)$ is true, then $p(k+1)$ is true.
We assumed $p(k)$ is true:
$$p(k):\text{ } \dfrac{1}{1\cdot 3}+\dfrac{1}{3\cdot 5}+\dfrac{1}{5\cdot 7}+\cdots+\dfrac{1}{(2k-1)(2k+1)}=\dfrac{k}{2k+1}.\tag1$$
We have to prove:
$$p(k+1):\text{ } \dfrac{1}{1\cdot 3}+\dfrac{1}{3\cdot 5}+\dfrac{1}{5\cdot 7}+\cdots+\dfrac{1}{(2k-1)(2k+1)}+\dfrac{1}{(2(k+1)-1)(2(k+1)+1)}=\dfrac{k+1}{2(k+1)+1}.$$
$$p(k+1):\text{ } \dfrac{1}{1\cdot 3}+\dfrac{1}{3\cdot 5}+\dfrac{1}{5\cdot 7}+\cdots+\dfrac{1}{(2k-1)(2k+1)}+\dfrac{1}{(2k+1)(2k+3)}=\dfrac{k+1}{2k+3}.\tag2$$
In Eq. $(1)$ add the term $\dfrac{1}{(2k+1)(2k+3)})$ on each side:
$$\begin{align*}
\dfrac{1}{1\cdot 3}+\dfrac{1}{3\cdot 5}+\dfrac{1}{5\cdot 7}+\cdots+\dfrac{1}{(2k-1)(2k+1)}+\dfrac{1}{(2k+1)(2k+3)}&=\dfrac{k}{2k+1}+\dfrac{1}{(2k+1)(2k+3)}\\
\dfrac{1}{1\cdot 3}+\dfrac{1}{3\cdot 5}+\dfrac{1}{5\cdot 7}+\cdots+\dfrac{1}{(2k-1)(2k+1)}+\dfrac{1}{(2k+1)(2k+3)}&=\dfrac{k(2k+3)+1}{(2k+1)(2k+3)}\\
\dfrac{1}{1\cdot 3}+\dfrac{1}{3\cdot 5}+\dfrac{1}{5\cdot 7}+\cdots+\dfrac{1}{(2k-1)(2k+1)}+\dfrac{1}{(2k+1)(2k+3)}&=\dfrac{2k^2+3k+1}{(2k+1)(2k+3)}\\
\dfrac{1}{1\cdot 3}+\dfrac{1}{3\cdot 5}+\dfrac{1}{5\cdot 7}+\cdots+\dfrac{1}{(2k-1)(2k+1)}+\dfrac{1}{(2k+1)(2k+3)}&=\dfrac{(2k+1)(k+1)}{(2k+1)(2k+3)}\\
\dfrac{1}{1\cdot 3}+\dfrac{1}{3\cdot 5}+\dfrac{1}{5\cdot 7}+\cdots+\dfrac{1}{(2k-1)(2k+1)}+\dfrac{1}{(2k+1)(2k+3)}=\dfrac{k+1}{2k+3}.
\end{align*}$$
We got Eq. $(2)$, therefore if $p(k)$ is true, then $p(k+1)$ is also true.
Both steps being proved, the given formula is true for any natural number $n$.
