Question:

$\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots+\frac{1}{(2 n+1)(2 n+3)}=\frac{n}{3(2 n+3)}$

Solution:

Let P(n) be the given statement.

Now,

$P(n)=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots+\frac{1}{(2 n+1)(2 n+3)}=\frac{n}{3(2 n+3)}$

Step 1;

$P(1)=\frac{1}{3.5}=\frac{1}{15}=\frac{1}{3(2+3)}$

Hence, $P(1)$ is true.

Step 2:

Let $P(m)$ be true.

Then,

$\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots+\frac{1}{(2 m+1)(2 m+3)}=\frac{m}{3(2 m+3)}$

To prove : $P(m+1)$ is true.

That is,

$\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots+\frac{1}{(2 m+3)(2 m+5)}=\frac{m+1}{3(2 m+5)}$

To prove : $P(m+1)$ is true.

That is,

$\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots+\frac{1}{(2 m+3)(2 m+5)}=\frac{m+1}{3(2 m+5)}$

Now,

$P(m)=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots+\frac{1}{(2 m+1)(2 m+3)}=\frac{m}{3(2 m+3)}$

$\Rightarrow \frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 m+1)(2 m+3)}+\frac{1}{(2 m+3)(2 m+5)}=\frac{m}{3(2 m+3)}+\frac{1}{(2 m+3)(2 m+5)}$

$\left[\right.$ Adding $\frac{1}{(2 m+3)(2 m+5)}$ to both side $\left.s\right]$

$\Rightarrow \frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 m+3)(2 m+5)}=\frac{2 m^{2}+5 m+3}{3(2 m+3)(2 m+5)}=\frac{(2 m+3)(m+1)}{3(2 m+3)(2 m+5)}=\frac{m+1}{3(2 m+5)}$

Thus, $P(m+1)$ is true.

By the principle of mathematical induction, $P(n)$ is true for all $n \in N$.