When has this sequence a triangular shape? Can someone explain to me why y[n] has a triangular shape? From what I have found, there is a specific range $n_0$ where y[n] is triangular.
But right now I have no clue where to start.

$^2$ Search in a table    of  Fourier transform   properties
 A: A simple way to see this: taking the square of a Fourier transform is just multiplying its Fourier transform with itself. Now, taking the inverse transform of a product of two Fourier transforms is the same as convolving the two original functions. Here you are convolving a step function with itself. Now, think about how convolutions work and it should be evident that the result is a triangle function. But if you don't see it, just work out the convolution product of two step functions.
EDIT: Here's a detailed worked out solution. Using the definition from wikipedia for the DFT:
$$\text{DFT}(x[n])=\sum_{n=0}^{N-1}(u[n]-u[n-n_0])e^{-\frac{2\pi i}{N}k n} = \sum_{n=0}^{n_0-1}e^{-\frac{2\pi i}{N}k n} \; .$$
If you multiply this DFT with itself, you get the following sum
$$(\text{DFT}(x[n]))^2= \left(\sum_{n=0}^{n_0-1}e^{-\frac{2\pi i}{N}k n}\right)\left(\sum_{m=0}^{n_0-1}e^{-\frac{2\pi i}{N}k m}\right) = \sum_{n=0}^{n_0-1}\sum_{m=0}^{n_0-1}e^{-\frac{2\pi i}{N}k (m+n)} \; .$$
With a change of variables $t=m+n$, you can now write this as
$$(\text{DFT}(x[n]))^2 = \sum_{t=0}^{2(n_0-1)} z[t] e^{-\frac{2\pi i}{N}k t} \; .$$
where $z[t]$ is 
$$z[t]=\begin{cases}1, \text{ if } t=0=0+0,\\
2, \text{ if } t=1=1+0=0+1,\\
3, \text{ if } t=2=2+0=1+1=0+2,\\
\ldots\\
2, \text{ if } t=2n_0-3=n_0-1+n-0-2=n_0-2+n_0-1,\\
1, \text{ if } t=2n_0-2=n_0-1+n_0-1,\\
\end{cases}$$
which is a triangular function. Note however that if $2(n_0-1)\geq N$, then because of periodicity, $z[t]\neq y[t]$. You'll have to add the $z[t]$'s corresponding to a same frequency to obtain the correct expression for $y[t]$, since $y[t]$ is only defined over the range $[0,N-1]$ just like $x[n]$. 
So the range they are talking about in the question is $0\leq2(n_0-1)<N$.
