puzzle answer: avg(sums(a set))
This is the answer to the last math puzzle. You should check out the puzzle before you read the answer!
It turns out that
avg(sums(S)) = ½ ∑S.
Why?
The main observation is that, for any T ⊂ S, avg({∑ T, ∑(S-T)}) = ½ ∑ S. We should also note that if ∑ T1 = ∑ T2, then ∑ S-T1 = ∑ S-T2; this means that whenever the sums of T1 and T2 overlap in sums(S), so do the sums of their complements. If we think of adding each pair {∑ T, ∑(S-T)} one at a time to build sums(S), then either both elements of the pair are already in the set, or both are not — either way, the average remains the same.