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Guess ???
1. x[n] oscillates between $ {1 \over 2} and {1 \over -2} $
2. x[0] = 10
3. x[1] = $ a_0 + {1 \over 2} $
4. x[n] has period 2
Solution
x[n] = 10 + $ sin[{\pi \over 2}n] + {1 \over 2}sin[\pi n] $
