Item — HN Reader

But can you not? Assume sum, 1..(N+1)/2..N

Have you not actually built the same proof via induction in both cases with one of them starting from the middle and subsequently including left and right terms at a unit away (you actually do it in reverse but the crux still holds)? Such that S[0] gives you (N+1)/2 and S[(N-1)/2] gives you the total sum.

The argument would be like S[i] = (2i-1)(N+1)/2 only you'd be proving it using induction i.e. given S[i-1], finding S[i].

All it ever matters for this problem is that to prove it as such, you somehow have to add up all of the numbers. The "different" proofs you presented are actually the same since for addition, the order of operation does not matter due to associative and commutative properties. A good question would be to see if any of the proofs still remain valid when either of these properties are removed from an operation.

You're handwaving, but I think there is a middle-ground in this proof:

Sum(i=1..n, i)

= Sum(i=1..n/2, i) + Sum(i=1..n/2, n+1-i)

= Sum(i=1..n/2, n+1)

I'm still interested in the general question, of whether some proofs have big gaps between them. The more complex the proofs, the more obvious this would be; my examples are unfortunately simple. Something like proving the fundamental theorem of algebra using Rouche's Theorem (complex analysis) vs. field theory. But I don't know enough math to compare those.