April 2nd, 2014, 03:16 PM  #11 
Senior Member Joined: May 2013 From: EspaĆ±a Posts: 151 Thanks: 4 
Hello. 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: ... Regards. 
April 2nd, 2014, 03:48 PM  #12 
Math Team Joined: Apr 2012 Posts: 1,579 Thanks: 22 
Great stuff. Just heard from my friend. I'll be doing this midMay or so!

April 2nd, 2014, 04:37 PM  #13 
Math Team Joined: Apr 2012 Posts: 1,579 Thanks: 22 
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.

April 2nd, 2014, 08:52 PM  #14 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 
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) 
April 3rd, 2014, 05:02 AM  #15 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
Maybe you can mine something from this: 

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