Take your final term as $$\frac{1}{(n-2)(n-1)(n)(n+1)}$$
If we try to write it in another form, we have $$\frac{1}{3}\left ( \frac{1}{(n-2)(n-1)(n)}-\frac{1}{(n-1)(n)(n+1)} \right )$$
OR
$$\frac{1}{3}\left ( X_{n}-X_{n+1}\right )$$
Can you proceed from here by taking summation ?