For n≥0 the terms are $2^nz^{-n}$, whose geometric series converges only for |z|>2.
For n<0 put m=−n>0. The terms become $2^mz^m$, which require |z|<1/2. No z satisfies both conditions.
Therefore the bilateral z-transform has **no ROC, D**. The two pole radii do not create an annulus because the required inequalities point away from each other.
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