Quote (Minkomonster @ 7 Feb 2014 14:46)
I explained it in terms of arithmetic series. And I specifically spoke to that. This is the first time you have given a name to the concept you deployed to solve the problem and you have still neglected to explain why it works. Further more, the math in your thread is jumbled and convoluted. To the outside reader it is indecipherable. Your solution is not simple for these reasons.
As an aside, I talked to OP in private messages and he validated my assertion by saying you left him more confused than when he asked the question. You didn't help shit. He wasn't even asking for a solution, he was asking for guidance on how to go about solving it. You failed to provide that. You failed.
you went and read my post, if the first line does not clearly indicate that it works by going down in a diagonal way
than i really have to wonder
if you call it jumbled and convoluted then it cannot be helped, the notation is straight forward
but just in case let me explain it in a more pictorial way:
if you mirror across the cross diagonal which has all n's in it
the two elements which then fall together add up to 2n (see the second line in the post)
which immediately gives [(n^2-n)/2]*2n = n^3-n^2 so one just has to add the value of the cross diagonal with is n time n
and the result is n^3
you can as well take the sum of the two 'mirrored' cells which add up to 2n and replace them each with the average namely n
what you have then is an nxn matrix with all n's in it, the sum of the elements is trivially n^3
as said, it is just a matter of recognising the structure/pattern