Would love to read a formal proof of this if one exists!
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The nth group of 1) is the range [½n(n-1)+1, ½n(n+1)], which sums to ½(n+n³). The nth group of 2) therefore sums to ½(2n-1 + (2n-1)³) = 4n³-6n²+4n-1. This just happens to be equal to n⁴-(n-1)⁴, which is what you need for induction to work.
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intuitively obvious!
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Why obvious?!
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For those wondering, here's a proof: each of the remaining groups contains 2n - 1 numbers, and the last number in the nth group is the sum of the first 2n - 1 numbers, or n(2n - 1) = 2n^2 - n. The first number is then n(2n - 1) - (2n - 1) + 1 = 2n^2 - 3n + 2. (1/)
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The average in each group is then 2n^2 - 2n + 1, and the total is just (2n - 1)(2n^2 - 2n + 1) = [n^2 - (n - 1)^2][n^2 + (n - 1)^2] = n^4 - (n - 1)^4. Everything falls into place nicely. (2/2)
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Awesome
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Very cool. Love this relationship.
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Very interesting. I think it should be revised so that "every second group" is listed as (2,3),(7,8,9,10),...
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