DT Signal:
1. Signal is periodic with N = 4
2. $ \sum_{k = 0}^{3}x[n] = (2 + j) $
3. $ a_{1} = a{2}\, $
4. for the given value of k, $ e^{jk\frac{2\pi}{N}} = -1\, $, then that $ a_{k} = \frac{1}{2}\, $
5. All other $ a_{k} = 0\, $
Solution
$ a_{0} = \frac{2 + j}{4} $
$ a_{1} = \frac{1}{2} $
$ a_{2} = \frac{1}{2} $
$ a_{3} = 0 $
$ x[n] = \frac{2 + j}{4} + \frac{1}{2}e^{j\frac{\pi}{2}n} + \frac{1}{2}e^{j\pi n} $
