# Prove the following by using the principle of mathematical induction for all $n \in N$  $\bigg( 1+ \large\frac{3}{1} \bigg)$$\bigg( 1+ \large\frac{5}{4} \bigg)$$\bigg( 1+ \large\frac{7}{9} \bigg)$$...\bigg( 1+ \large\frac{(2n+1)}{n^2} \bigg)$$=(n+1)^2$

Let the given statement be $P(n)$, i.e.,
$P(n) : \bigg( 1+ \large\frac{3}{1} \bigg)$$\bigg( 1+ \large\frac{5}{4} \bigg)$$\bigg( 1+ \large\frac{7}{9} \bigg)$$...\bigg( 1+ \large\frac{(2n+1)}{n^2} \bigg)$$=(n+1)^2$
For $n = 1$, we have
$P(1) : \bigg( 1+ \large\frac{3}{1} \bigg)$$=4=(1+1)^2=2^2=4, which is true. Let P(k) be true for some positive integer k, i.e., \bigg( 1+ \large\frac{3}{1} \bigg)$$\bigg( 1+ \large\frac{5}{4} \bigg)$$\bigg( 1+ \large\frac{7}{9} \bigg)$$...\bigg( 1+ \large\frac{(2k+1)}{k^2} \bigg)$$=(k+1)^2------------(i) We shall now prove that P(k+1) is true. Consider \bigg[ \bigg( 1+ \large\frac{3}{1} \bigg)$$\bigg( 1+ \large\frac{5}{4} \bigg)$$\bigg( 1+ \large\frac{7}{9} \bigg)$$ ...\bigg(1+ \large\frac{(2k+1)}{k^2} \bigg)$$\bigg]$$ \bigg\{ 1+ \large\frac{ \{2(k+1)+1\}}{(k+1)^2} \bigg\}$
$=(k+1)^2 \bigg[ 1+ \large\frac{ 2(k+1)+1}{(k+1)^2} \bigg]\qquad$ [ Using (i) ]
$=(k+1)^2 \bigg[ \large\frac{ (k+1)^2+2(k+1)+1}{(k+1)^2} \bigg]$
$= (k+1)^2+2(k+1)+1$
$= \{ (k+1)+1 \}^2$
Thus, $P(k+1)$ is true whenever $P(k)$ is true.
Hence, by the principle of mathematical induction, statement $P(n)$ is true for all natural numbers i.e., $n$.