Answer
See below.
Work Step by Step
1. Test for $n=1$, we have $LHS=\frac{1}{1\cdot3}=\frac{1}{3}$ and $RHS=\frac{1}{2(1)+1}=\frac{1}{3}$, thus it is true.
2. Assume the formula is true for $n=k$, we have
$\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{(2k-1)(2k+1)}=\frac{k}{2k+1}$
3. For $n=k+1$, we have $LHS=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{(2k-1)(2k+1)}+\frac{1}{(2(k+1)-1)(2(k+1)+1)}=\frac{k}{2k+1}+\frac{1}{(2k+2-1)(2k+2+1)}=\frac{k}{2k+1}+\frac{1}{(2k+1)(2k+3)}=\frac{1}{2k+1}(k+\frac{1}{2k+3})=\frac{1}{2k+1}(\frac{2k^2+3k+1}{2k+3})=\frac{1}{2k+1}(\frac{(2k+1)(k+1)}{2k+3})=\frac{k+1}{2(k+1)+1}=RHS$
4. We conclude that the formula is true for any $n$.
