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!

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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.