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.
 A: For $n=1$ you have $3n-2=1$ and there is only one number - the sum is $1$
For $n=2$ you have $3n-2=4$ and the sum is $2+3+4=9$
For $n=3$ you have $3n-2=7$ and the sum is $3+4+5+6+7=25$
The numbers $n$ and $3n-2$ are part of the recipe by which the sum is put together. There is no problem using $3n-2$ in such a recipe, provided it is consistent. Writing out the first few as I have done sometimes helps to see what is happening. There are often little peculiarities with the first term (here the sum consists of one item), so the second and third can give a better sense of what happens in general.
The next thing to do is to try to write down what happens in general to make the conclusion obvious. Then you have to prove it always works.


Since there is a full solution posted, here is a different suggestion. With $2n-1=r$ the sum is $(r-(n-1)) + (r-(n-2))+\dots +(r-1)+r+(r+1)+\dots +(r+(n-1))$. There are $2n-1=r$ terms and pairing them from the outside, or taking the average term and multiplying by the number of terms, gives a sum of $r^2$

A: Using the identity,
$$1+2+\cdots+n=\frac{n(n+1)}2$$
we get
$$\begin{align}
n+(n+1)+\cdots+3n-2 & =(1+2+\cdots+3n-2)-(1+2+\cdots+(n-1)\\
& = \frac{(3n-2)(3n-1)}2-\frac{n(n-1)}2\\
& = \frac{8n^2-8n+2}2\\
& = 4n^2-4n+1\\
&= (2n-1)^2.
\end{align}$$
I'm willing to bet there's a typo in your book.
A: Yes, It will be $(2n-1)^2$
The proof as follows, $$n+(n+1)+(n+2)+\dots+[n+(2n-2)]\\=\underbrace{(n+n+\dots+n)}_{(2n-2+1)=2n-1 \text{ times}}+\{1+2+\dots+(2n-2)\}\\=(2n-1)n+\frac {(2n-2)(2n-1)}2\\=(2n-1)(n+n-1)=(2n-1)^2$$
Everyone's answer posted above are extraordinary, I am not intended to undermine them. I posted this because, it was a slightly different approach.
A: Let 
$s_n
=\sum_{k=n}^{3n-2} k 
$.
$s_1 = 1$.
$s_2 = 2+3+4 = 9$.
$s_3 = 3+4+5+6+7 = 25$.
Conjecture:
$s_n = (2n-1)^2$.
$s_{n+1}-s_n
=\sum_{k=n+1}^{3n+1} k-\sum_{k=n}^{3n-2} k
=(3n-1+3n+3n+1)-(n)
=8n
$.
If
$s_n = (2n-1)^2$,
$s_{n+1}
=(2n=1)^2+8n
=4n^2-4n+1+8n
=4n^2+4n+1
=(2n+1)^2
$.
This prove the result
by induction.
A: 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 $$S_n = 4n^2 - 4n + 1 = (2n-1)^2$$
