Question

Use mathematical induction to prove:
$\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\dots +\frac{1}{\left(2n-1\right)\left(2n+1\right)}=\frac{n}{2n+1}$

Answer 1

$\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\cdots +\frac{1}{(2n-1)(2n+1)}=\frac{n}{2n+1}$

$p\left(1\right)=\frac{1}{1.3}=\frac{1}{3}$ true

$p\left(n\right)$ is true for some $k\in N$

i.e $\frac{1}{1.3}+\frac{1}{3.5}+\cdots +\frac{1}{(2k-1)(2k+1)}=\frac{k}{2k+1}$

we need to prove $P(k+1)$ is true whenever $p\left(k\right)$ is true.

Now $\frac{1}{1.3}+\frac{1}{3.5}+\cdots +\frac{1}{(2k-1)(2k+1)}+\frac{1}{(2k+1)(2k+3)}$

$=[\frac{1}{1.3}+\frac{1}{3.5}+\cdots +\frac{1}{(2k-1)(2k+1)}]+\frac{1}{(2k+1)(2k+3)}$

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

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

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

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

Thus $p(k+1)$ is true.

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Thus PMI $p\left(n\right)$ is true $\forall n\in N$