Prove that for each number $n \in \mathbb N$ is the sum of numbers $n, n + 1, n + 2, ..., 3n - 2$ equal to the second power of a natural number. I've got a homework from maths:

Prove that for each number $n \in \mathbb N$ is the sum of numbers 
  $n, n + 1, n + 2, ..., 3n - 2$ equal to the second power of a natural
  number.

I don't actually get the task. I would assume that the sequence would continue like this: $n, n + 1, n + 2, n + 3, ..., n + k$, but not $3n - 2$. Isn't there something wrong with the book?
The key of the book says only:
$$ S_n = (2n-2)^2 $$
where $ S_n$ is the sum of all the numbers.
Thank You for Your reply,
Dominik

EDIT
Sorry, I made a mistake. It really is
$$ S_n = (2n-1)^2 $$
I am really sorry.

By the trick that we know from Gauss we get that $$\sum_{i = 1}^s i = \frac{s(s+1)}{2}$$
If we fill in $s = 3n - 2$ and substract the sum up to $s = n-1$ we see that $$S_n = \sum_{i = n}^{3n - 2} i = \sum_{i = 1}^{3n - 2} i - \sum_{i = 1}^{n-1} i$$
so $$S_n =  \frac{(3n - 2)(3n - 1)}{2} - \frac{(n-1)(n)}{2}$$ which gives us $$S_n = 4n^2 - 4n + 1 = (2n-1)^2$$
