Finding the discrete time Fourier series for signal I think I did everything correctly here, but I must be missing something still.
We have the following signal: 
My approach:
We are told that the signal has period $N = 4$
We know $$Y[k] = \frac{1}{N}\sum_{n = 0}^{3}y[n]e^{-j\frac{\pi}{2}nk}$$ The signal only has two non-zero terms at $n \mod 4 = 0$ and $n \mod 4 = 2$
So, $$Y[k] = \frac{1}{N}\left(e^{-j\frac{\pi}{2}k} -\frac{1}{2} e^{-j \pi k}\right)$$
$$Y[k] =  \frac{1}{4}\left( (-j)^{k} - \frac{1}{2} (-1)^{k}\right)$$
$$Y[k] =  \begin{cases}
 \frac{1}{8} &amp;\text{if}\, k = 0\\
 \frac{1}{4}(.5 - j)&amp;\text{if}\, k = 1\\
\frac{-3}{8} &amp;\text{if}\, k = 2\\
 \frac{1}{4}(j + .5) &amp;\text{otherwise}
\end{cases}$$
However the reported answer is actually
$$Y[k] =  \begin{cases}
 \frac{1}{8} &amp;\text{if}\, k = 0\\
 \frac{3}{8} &amp;\text{if}\, k = 1\\
\frac{1}{8} &amp;\text{if}\, k = 2\\
 \frac{3}{8} &amp;\, k = 3
\end{cases}$$
Where am I going wrong? Any help is very much appreciated :)
Your first term in the second equation is wrong for $n=0$. You put in the term for $n=1$ . It should be
$$Y[k] = \frac{1}{N}\left(1 -\frac{1}{2} e^{-j \pi k}\right)$$