Using the principle of mathematical induction, prove each of the following for all n ϵ N:
$\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \ldots\left\{1+\frac{1}{n}\right\}=(n+1)$
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.