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

To Prove:

$\frac{1}{1 \times 3}+\frac{1}{(3 \times 5)}+\ldots \ldots+\frac{1}{(2 n-1) \times(2 n+1)}=\frac{n}{(2 n+1)}$

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

Let $\mathrm{P}(\mathrm{n}): \frac{1}{1 \times 3}+\frac{1}{(3 \times 5)}+\ldots \ldots+\frac{1}{(2 n-1) \times(2 n+1)}=\frac{n}{(2 n+1)}$

For n = 1

$\mathrm{LHS}=\frac{1}{1 \times 3}=\frac{1}{3}$

$\mathrm{RHS}=\frac{1}{(2+1)}=\frac{1}{3}$

Hence, LHS = RHS

P(n) is true for n = 1

Assume P(k) is true

$=\frac{1}{1 \times 3}+\frac{1}{(3 \times 5)}+\ldots \ldots+\frac{1}{(2 k-1) \times(2 k+1)}=\frac{k}{(2 k+1)} \ldots \ldots(1)$

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

$\mathrm{RHS}=\frac{k+1}{(2(k+1)+1)}=\frac{k+1}{(2 k+3)}$

$\mathrm{LHS}=\frac{1}{1 \times 3}+\frac{1}{(3 \times 5)}+\ldots \ldots+\frac{1}{(2(k+1)-1) \times(2(k+1)+1)}$

$=\frac{1}{1 \times 3}+\frac{1}{(3 \times 5)}+\ldots \ldots+\frac{1}{(2 k-1) \times(2 k+1)}+\frac{1}{(2 k+1) \times(2 k+3)}$

[Writing the second last term]

$=\frac{k}{(2 k+1)}+\frac{1}{(2 k+1) \times(2 k+3)}[$ Using 1$]$

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

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

$=\frac{k+1}{(2 k+3)}$

(Splitting the numerator and cancelling the common factor)

= RHS

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

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

By the principle of mathematical induction, $P(n)$ is true for $x$

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

Hence proved.