= Solution
The <base case> $n=3$ is
$$
\frac1{3(3-2)}=\frac13=\frac34-\frac5{12}.
$$
Assume the displayed formula holds at some $n\geq3$. Adding the next term yields
$$
\begin{aligned}
\sum_{i=3}^{n+1}\frac1{i(i-2)}
&=\frac34-\frac{2n-1}{2n(n-1)}+\frac1{(n+1)(n-1)}\\
&=\frac34-\frac{(2n-1)(n+1)-2n}{2n(n-1)(n+1)}\\
&=\frac34-\frac{2n+1}{2n(n+1)}.
\end{aligned}
$$
The last step uses $(2n-1)(n+1)-2n=(2n+1)(n-1)$. This is the required formula at $n+1$, completing the proof by <mathematical induction>. \b[The identity holds for all $n\geq3$.]
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