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April 2nd, 2014, 03:16 PM   #11
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Let "n"=number of consecutive addends, such that:


In addition, the difference of the first Addend of a succession, with the end of the previous one, is a number squared:


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April 2nd, 2014, 03:48 PM   #12
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Great stuff. Just heard from my friend. I'll be doing this mid-May or so!
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April 2nd, 2014, 04:37 PM   #13
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Just glancing over the material here a bit more closely and I can see that Soroban's first demo is particularly germane, as it basically is an alternative view of the fact that the difference between the squares of the n+1st and nth triangular numbers is the cube of n+1.
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April 2nd, 2014, 08:52 PM   #14
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As a "break", give them this li'l "match moving" puzzle:

II = VI : move 1 match only to end up with both sides equal

I = VT : right side is now SQRT(1)
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April 3rd, 2014, 05:02 AM   #15
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Maybe you can mine something from this:
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