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?
 A: Write $\frac{1}{(3n-2)(3n+1)}=\frac{1}{3}(\frac{1}{3n-2}-\frac{1}{3n+1})$ before summing.
A: This is as nice task for induction, isn't it?
for $n=1$ we clearly have
$$\sum_{i=1}^1 \frac{1}{(3k-2)(3k+1)} = \frac{1}{4} = \frac{1}{3\cdot 1+1}$$
The induction step is not so difficult as well - it was 2 lines on my paper.
A: To find the sum of $n$ terms of a series each term of which is composed of the reciprocal of the product of $r$ factors in arithmetical progression, the first factors of the several terms being in the same arithmetical progression, use following
Write down the $n^{th}$ term, strike off a factor from the beginning, divide by the number of factors so diminished and by the common difference, change the sign and add a constant
In the example given, $(1,4),(4,7),(7,10)$ are in A.P. difference being $\color{red}3$ between each number of any pair also $1, 4, 7$ are in A.P with the same difference. 
$$\therefore Sum=\frac{-1}{\color{red}3\times (3n+1)}+c$$
When $n=1$    Sum=$\frac{1}{4} \Rightarrow c=\frac{1}{4}$
$$\therefore Sum=\frac{-1}{3\times (3n+1)}+\frac{1}{4}=\frac{n}{3n+1}$$
A: 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.
