Prove the following by using the principle of mathematical induction for all $n \in N$:
$\frac{1}{3 \times 5} + \frac{1}{5 \times 7} + \frac{1}{7 \times 9} + \ldots + \frac{1}{(2n+1)(2n+3)} = \frac{n}{3(2n+3)}$

Let the given statement be $P(n)$:
$P(n): \frac{1}{3 \times 5} + \frac{1}{5 \times 7} + \frac{1}{7 \times 9} + \ldots + \frac{1}{(2n+1)(2n+3)} = \frac{n}{3(2n+3)}$
For $n=1$,we have:
$P(1): \frac{1}{3 \times 5} = \frac{1}{3(2(1)+3)} = \frac{1}{3 \times 5}$,which is true.
Assume $P(k)$ is true for some positive integer $k$,i.e.,
$P(k): \frac{1}{3 \times 5} + \frac{1}{5 \times 7} + \ldots + \frac{1}{(2k+1)(2k+3)} = \frac{k}{3(2k+3)}$
We shall now prove that $P(k+1)$ is true.
Consider the sum up to $(k+1)$ terms:
$\left[\frac{1}{3 \times 5} + \ldots + \frac{1}{(2k+1)(2k+3)}\right] + \frac{1}{(2(k+1)+1)(2(k+1)+3)}$
$= \frac{k}{3(2k+3)} + \frac{1}{(2k+3)(2k+5)}$
$= \frac{1}{2k+3} \left[ \frac{k}{3} + \frac{1}{2k+5} \right]$
$= \frac{1}{2k+3} \left[ \frac{k(2k+5) + 3}{3(2k+5)} \right]$
$= \frac{2k^2 + 5k + 3}{3(2k+3)(2k+5)}$
$= \frac{(k+1)(2k+3)}{3(2k+3)(2k+5)}$
$= \frac{k+1}{3(2k+5)} = \frac{k+1}{3(2(k+1)+3)}$
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,the statement $P(n)$ is true for all $n \in N$.