Prove the following by using the principle of mathematical induction for all n ∈ N:

Solution:

Let the given statement be P(n), i.e.,

$\mathrm{P}(n):\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \ldots\left(1+\frac{(2 n+1)}{n^{2}}\right)=(n+1)^{2}$

For n = 1, we have

$P(1):\left(1+\frac{3}{1}\right)=4=(1+1)^{2}=2^{2}=4$, which is true.

Let P(k) be true for some positive integer k, i.e.,

$\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \ldots\left(1+\frac{(2 k+1)}{k^{2}}\right)=(k+1)^{2}$ $\ldots(1)$

We shall now prove that P(k + 1) is true.

Consider

$\left[\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \ldots\left(1+\frac{(2 k+1)}{k^{2}}\right)\right]\left\{1+\frac{\{2(k+1)+1\}}{(k+1)^{2}}\right\}$

$=(k+1)^{2}\left(1+\frac{2(k+1)+1}{(k+1)^{2}}\right)$ [Using(1)]

$=(k+1)^{2}\left[\frac{(k+1)^{2}+2(k+1)+1}{(k+1)^{2}}\right]$

$=(k+1)^{2}+2(k+1)+1$

$=\{(k+1)+1\}^{2}$

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.