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:

$\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \ldots\left\{1+\frac{1}{n}\right\}=(n+1)$

Solution:

To Prove:

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

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

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

For n = 1

$\mathrm{LHS}=1+\frac{1}{1}=2$

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

Hence, LHS = RHS

P(n) is true for n = 1

Assume P(k) is true

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

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

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

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

[Now writing the second last term]

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

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

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

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

$=\mathrm{k}+2$

$=\mathrm{RHS}$

$\mathrm{LHS}=\mathrm{RHS}$

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.