Answer
See explanations.
Work Step by Step
Step 1. Prove that the formula is true for $n=1$: $LHS=\frac{1}{3}$, $RHS=\frac{1}{2+1}=\frac{1}{3}$, thus it is true for $n=1$
Step 2. Assume the formula is true for $n=k$: we have $\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{(2k-1)(2k+1)}=\frac{k}{2k+1}$
Step 3. Prove that it is also true for $n=k+1$:
$LHS=\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{(2k-1)(2k+1)}+\frac{1}{(2k+1)(2k+3)}
=\frac{k}{2k+1}+\frac{1}{(2k+1)(2k+3)}=\frac{k(2k+3)+1}{(2k+1)(2k+3)}=\frac{2k^2+3k+1}{(2k+1)(2k+3)}
=\frac{(k+1)(2k+1)}{(2k+1)(2k+3)}=\frac{k+1}{2k+3}$
$RHS=\frac{k+1}{2k+3}=LHS$
Thus, the formula is also true for $n=k+1$
Step 4. With mathematical induction, we have proved that the formula is true for all natural numbers n.
