# Using the principle of mathematical induction, prove each of the following

Question:

Using the principle of mathematical induction, prove each of the following for all n ϵ N:

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

Solution:

To Prove:

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

Let us prove this question by principle of mathematical induction (PMI)

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

For n = 1

LHS $=1+\frac{3}{1}=4$

$\mathrm{RHS}=(1+1)^{2}=4$

Hence, LHS = RHS

$P(n)$ is true for $n=1$

Assume $P(k)$ is true

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

We will prove that P(k + 1) is true

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

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

[Now writing the second last term]

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

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

$=(k+1)^{2} \times\left\{1+\frac{(2 k+3)}{(k+1)^{2}}\right\}$

$=(k+1)^{2} \times\left\{\frac{(k+1)^{2}+(2 k+3)}{(k+1)^{2}}\right\}$

$=(k+1)^{2}+(2 k+3)$

$=k^{2}+2 k+1+2 k+3$

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

= RHS

LHS = RHS

Therefore, $P(k+1)$ is true whenever $P(k)$ is true

By the principle of mathematical induction, $\mathrm{P}(\mathrm{n})$ is true for

Where $\mathrm{n}$ is a natural number

Hence proved.