= Solution
{c}
In level $n$, each group is
$$
2^{-(n+1)}+2^{-(n+1)}-2^{-n}=0,
$$
and there are $2^n$ such groups. The <partial sums> at completed groups are therefore zero. Inside a group of level $n$, the only other <partial sums> are $2^{-(n+1)}$ and $2^{-n}$. Every level contains finitely many terms, so the level containing the $N$th term tends to infinity with $N$. Thus all <partial sums> lie between zero and a bound tending to zero. By the <squeeze theorem>, the <series> has sum $\boxed{0}$.
For the next two parts, label the positive occurrences of size $2^{-(n+1)}$ by $P_n$ and the negative occurrences of size $-2^{-n}$ by $N_n$. There are exactly $2^{n+1}$ occurrences in $P_n$ and $2^n$ in $N_n$. Occurrences with equal value retain separate labels, so checking their counts verifies a <bijection> of terms.
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