Prove the following by using the principle of mathematical induction for all $n \in N$:
$\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \dots \left(1+\frac{2n+1}{n^{2}}\right)=(n+1)^{2}$

Let the given statement be $P(n)$,i.e.,
$P(n): \left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \dots \left(1+\frac{2n+1}{n^{2}}\right)=(n+1)^{2}$
For $n=1$,we have
$P(1): \left(1+\frac{3}{1}\right) = 4 = (1+1)^{2} = 2^{2} = 4$,which is true.
Assume $P(k)$ is true for some positive integer $k$,i.e.,
$\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \dots \left(1+\frac{2k+1}{k^{2}}\right)=(k+1)^{2}$ $(i)$
We shall now prove that $P(k+1)$ is true.
Consider the product up to $(k+1)$ terms:
$\left[\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right) \dots \left(1+\frac{2k+1}{k^{2}}\right)\right] \left\{1+\frac{2(k+1)+1}{(k+1)^{2}}\right\}$
$= (k+1)^{2} \left(1+\frac{2k+3}{(k+1)^{2}}\right)$ [Using $(i)$]
$= (k+1)^{2} \left[\frac{(k+1)^{2} + 2k + 3}{(k+1)^{2}}\right]$
$= (k+1)^{2} + 2k + 3$
$= k^{2} + 2k + 1 + 2k + 3 = k^{2} + 4k + 4$
$= (k+2)^{2} = \{(k+1)+1\}^{2}$
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 $n \in N$.