We are given the following information about a signal x[n]:
1. x[n] is periodic with period 3
2. $ \sum^{2}_{n = 0} x[n] = 5 $
3. $ \sum^{2}_{n = 0} (-1)^{n} x[n] = 15 $
4. x[n] has minimum power among all signals that satisfy 1,2,3.
1. => $ x[n] = \frac{1}{3} \sum^{2}_{n = 0} a_k e^{j k \frac{2\pi}{3} n} $
2. => $ x[n] = \frac{5}{3} = a_0 $
3. => $ a_1 = \frac{1}{3}(15) = 5 $
$ a_2 = \frac{-1}{3}(15) = -5 $
Solution = $ \frac{5}{3} + 5e^{j \frac{2\pi}{3} t} - 5e^{j \frac{2\pi}{3} t} $
