Question

Prove the following by using the principle of mathematical induction for all $n\in N.$
$\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+\cdots +\frac{1}{(2n+1)(2n+3)}=\frac{n}{3(2n+3)}$

Answer 1

Step $\left(1\right):$ Assume given statement
Let the given statement be $P\left(n\right)$, i.e.,
$P\left(n\right)=\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+\cdots +\frac{1}{(2n+1)(2n+3)}=\frac{n}{3(2n+3)}$

Step $\left(2\right):$ Checking statement $P\left(n\right)$ for $n=1$
Put $n=1$ in $P\left(n\right)$, we get
$P\left(1\right):\frac{1}{3\cdot 5}=\frac{1}{3(2\cdot 1+3)}$
$\Rightarrow \frac{1}{15}=\frac{1}{15}$
Thus $P\left(n\right)$ is true for $n=1$

Step $\left(3\right):$ $P\left(n\right)$ for $n=K$.
Put $n=K$ in $P\left(n\right)$ and assume this is true for some natural number $K$ i.e.,
$P\left(k\right):\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+\cdots +\frac{1}{(2K+1)(2K+3)}=\frac{K}{3(2K+3)}\cdots \left(1\right)$

Step $\left(4\right):$ Checking statement $P\left(n\right)$ for $n=K+1$
Now,we shall prove that $P(K+1)$ is true whenever $P\left(K\right)$ is true.
Now we have
$\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+\cdots +\frac{1}{(2K+1)(2K+3)}+\frac{1}{\left\{2\right(K+1)+1\}\left\{2\right(K+1)+3\}}$
$=\frac{K}{3(2K+3)}+\frac{1}{(2K+3)(2K+5)}$ (using $\left(1\right)$)
$=\frac{1}{(2K+3)}\{\frac{K}{3}+\frac{1}{2K+5}\}$
$=\frac{1}{(2K+3)}\left\{\frac{2K^2+5K+3}{3(2K+5)}\right\}$
$=\frac{1}{(2K+3)}\left\{\frac{2K^2+2K+3K+3}{3(2K+5)}\right\}$
$=\frac{1}{(2K+3)}\left\{\frac{2K(K+1)+3(K+1)}{3(2K+5)}\right\}$
$=\frac{1}{2K+3}\left\{\frac{(K+1)(2K+3)}{3(2K+5)}\right\}$
$=\frac{K+1}{3(2K+5)}$

We can write it as $\frac{(K+1)}{3\left\{2\right(K+1)+3\}}$
Thus, $P(K+1)$ is true whenever $P\left(K\right)$ is true.

Final answer :
Therefore, by the principle of mathematical induction, statement $P\left(n\right)$ is true for all $n\in N$.