Answer
$\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\cdots+\frac{1}{(2n-1)\cdot(2n+1)}=\frac{n}{2n+1}$
Work Step by Step
Based on the existing pattern, we can write a general formula as:
$\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\cdots+\frac{1}{(2n-1)\cdot(2n+1)}=\frac{n}{2n+1}$
Proof: Let $P(n)$ be the above statement.
1. For $n=1$, we have $LHS=\frac{1}{1\cdot3}=\frac{1}{3}$ and $RHS=\frac{1}{2(1)+1}=\frac{1}{3}=LHS$, thus it is true for $P(1)$.
2. Assume it is true for $P(k)$, that is:
$\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\cdots+\frac{1}{(2k-1)\cdot(2k+1)}=\frac{k}{2k+1}$
3. For $n=k+1$, we have:
$LHS=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\cdots+\frac{1}{(2k-1)\cdot(2k+1)}+\frac{1}{(2k+1)\cdot(2k+3)}\\
=\frac{k}{2k+1}+\frac{1}{(2k+1)\cdot(2k+3)}=\frac{2k^2+3k+1}{(2k+1)\cdot(2k+3)}=\frac{(2k+1)(k+1)}{(2k+1)\cdot(2k+3)}=\frac{k+1}{2k+3}=RHS$
4. Thus $P(k+1)$ is also true and we proved the statement by mathematical induction.