joehx
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Posts posted by joehx
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What about
This could be split into 50 pairs, each given by (2n)^2 - (2n-1)^2 = 4n-1. Summing this between 1 and 50 gives 2n(n+1)-n = 5050. Not an "aha" solution, but I had to write it down
An alternating squares "aha" puzzle
in New Logic/Math Puzzles
Posted
I'm not sure it makes it any easier but you could pair from opposite ends, like in the sum to n.