Telescopic summation I'm working on this question:

Rewrite the following summation using sigma notation and then compute it 
  using the technique of telescoping summation.
  $$\frac{1}{2*5}+\frac{1}{3*6}+\frac{1}{4*7}+...+\frac{1}{(n-2)(n+1)}+\frac{1}{(n-1)(n+2)}  $$

My work:
I replaced the $n's$ for $i's$ and added the Sigma to it $$\sum_{i=1}^n \frac{1}{(i-2)(i+1)}+\frac{1}{(i-1)(i+2)}  $$
would it be wrong to replace every $i$ with $\frac{n(n+1)}{2}$? And then simplify the resulting equation. 
 A: HINT: Your sigma expression is not correct (you can see this, as by plugging in $i=0$ the second term in the expression is undefined). Consider that in the sum
$$
\frac{1}{2*5}+\frac{1}{3*6}+\frac{1}{4*7}+...+\frac{1}{(n-2)(n+1)}+\frac{1}{(n-1)(n+2)}
$$
every term is of the form $\frac{1}{(n-1)(n+2)}$ for some $n$. This should let you figure out what the initial index value should be in the sum [Hint: Compare $\frac{1}{2 \cdot 5} = \frac{1}{(i-1)(i+2)}$].
Since I noticed in the comments below that the poster is still unsure, you should expect to arrive at the sum 
$$
\frac{1}{2*5} + \cdots + \frac{1}{(n-1)(n+2)}
=\sum_{i = 3} ^n \frac{1}{(i-1)(i+2)}
$$
From this, similar to the method in @lab you can write this in a clearly telescoping form, using that $\frac{1}{i-1} - \frac{1}{i+2} = \frac{i+2 - (i-1)}{(i-1)(i+2)} = 3 \frac{1}{(i-1)(i+2)}$. Hence, 
$$
\sum_{i = 3} ^n \frac{1}{(i-1)(i+2)} = \sum_{i = 3} ^n \frac{1}{3} \left( \frac{1}{i-1} - \frac{1}{i+2} \right) = \frac{1}{3} \sum_{i = 3} ^n \left( \frac{1}{i-1} - \frac{1}{i+2} \right). 
$$
Perhaps you can take it from here?
A: HINT:
$$\frac1{(i-2)(i+1)}=\frac13\cdot\frac{i+1-(i-2)}{(i-2)(i+1)}=\frac13\left(\frac1{i-2}-\frac1{i+1}\right)$$
Similarly, $$\frac1{(i-1)(i+2)}=\cdots=\frac13\left(\frac1{i-1}-\frac1{i+2}\right)$$
Set $i=1,2,\cdots,n-1,n$ find the surviving terms 
