<s>[INST] How to get the given equality? I have the following sum ($n\in \Bbb N)$:
$$ \frac {1}{1 \times 4} + \frac {1}{4 \times 7} + \frac {1}{7 \times 10} +...+ \frac {1}{(3n - 2)(3n + 1)} \tag{1} $$
It can be proved that the sum is equal to 
$$ \frac{n}{3n + 1} \tag{2}$$
My question is, how do I get the equality? I mean, if I hadn't knew the formula $(2)$, how would I derive it?
 [/INST] Hint:
Use partial fraction of the ratio
$\frac{1}{(3n-2)(3n+1)}=\frac{1}{3}(\frac{1}{3n-2}-\frac{1}{3n+1})$
You will see mass cancellation such as below:
$\frac{1}{3}\left[1-\frac{1}{4}\right]$
$\frac{1}{3}\left[\frac{1}{4}-\frac{1}{7}\right]$
..
$\frac{1}{(3n-2)(3n+1)}=\frac{1}{3}(\frac{1}{3n-2}-\frac{1}{3n+1})$
Summing all you get $\left[\frac{1}{3}[1-(\frac{1}{3n+1})\right]$
Simplifying you get the result.
</s>