Prove the following by using the principle of mathematical induction for all $n \in N:$
$\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \ldots + \frac{1}{(2n-1)(2n+1)} = \frac{n}{2n+1}$

(N/A) Let $P(n)$ be the statement: $\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \ldots + \frac{1}{(2n-1)(2n+1)} = \frac{n}{2n+1}$.
Step $1$: For $n = 1$,$LHS$ $= \frac{1}{1 \cdot 3} = \frac{1}{3}$ and $RHS$ $= \frac{1}{2(1)+1} = \frac{1}{3}$. Since $LHS$ $=$ $RHS$,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \in N$,i.e.,$\frac{1}{1 \cdot 3} + \ldots + \frac{1}{(2k-1)(2k+1)} = \frac{k}{2k+1}$.
Step $3$: For $n = k+1$,we need to show $P(k+1)$ is true,i.e.,$\frac{1}{1 \cdot 3} + \ldots + \frac{1}{(2k+1)(2k+3)} = \frac{k+1}{2k+3}$.
$LHS$ $= \left(\frac{1}{1 \cdot 3} + \ldots + \frac{1}{(2k-1)(2k+1)}\right) + \frac{1}{(2k+1)(2k+3)}$
$= \frac{k}{2k+1} + \frac{1}{(2k+1)(2k+3)}$
$= \frac{k(2k+3) + 1}{(2k+1)(2k+3)} = \frac{2k^2 + 3k + 1}{(2k+1)(2k+3)}$
$= \frac{(2k+1)(k+1)}{(2k+1)(2k+3)} = \frac{k+1}{2k+3} = \text{RHS}$.
Thus,$P(k+1)$ is true whenever $P(k)$ is true. By the principle of mathematical induction,$P(n)$ is true for all $n \in N$.