Question

Prove the following by using principle of mathematical induction for all $n\in N:\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{(2n+1)(2n+3)}=\frac{n}{3(2n+3)}$

Answer 1

Let the given statement be $P\left(n\right)$ i.e.,
$P\left(n\right):\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{(2n+1)(2n+3)}=\frac{n}{3(2n+3)}$
For $n=1$ we have
$P\left(1\right):\frac{1}{3.5}=\frac{1}{3(2.1+3)}=\frac{1}{3.5}$, which is true.
Let $P\left(k\right)$ be true for some $k\in N$, i.e.,
$P\left(k\right):\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{(2k+1)(2k+3)}=\frac{k}{3(2k+3)}........\left(i\right)$
We shall now prove that $P(k+1)$ is true.
Consider
$\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{(2k+1)(2k+3)}+\frac{1}{\left\{2\right(k+1)+1\}\left\{2\right(k+1)+3\}}$
$=\frac{k}{3(2k+3)}+\frac{1}{(2k+3)(2k+5)}$ [Using $\left(i\right)$]
$=\frac{1}{(2k+3)}\{\frac{k}{3}+\frac{1}{(2k+5)}\}$
$=\frac{1}{(2k+3)}\left\{\frac{k(2k+5)+3}{3(2k+5)}\right\}$
$=\frac{1}{(2k+3)}\left\{\frac{2k^2+5k+3}{3(2k+5)}\right\}$
$=\frac{1}{(2k+3)}\left\{\frac{2k^2+2k+3k+3}{3(2k+5)}\right\}$
$=\frac{1}{(2k+3)}\left\{\frac{2k(k+1)+3(k+1)}{3(2k+5)}\right\}$
$=\frac{(k+1)(2k+3)}{3(2k+3)(2k+5)}$
$=\frac{(k+1)}{3\left\{2\right(k+1)+1\}}$
Thus $P(k+1)$ is true whenever $P\left(k\right)$ is true.
Hence, by the principle of mathematical induction statement $P\left(n\right)$ is true for all natural numbers i.e., $n$.