I'm not sure that I agree with your working so far - that is, if I'm understanding your requirements.
Take a look at the diagram I've attached.
I agree with this. This sum can be thought of as:
Originally Posted by rainer
(1) The sum of the lengths of the lines lying in the plane ; PLUSThis last sum, (3), can be found by taking the sum of the distances of one of the points in plane #1 - say, - from each of the points in plane #2, and multiplying by 4 (since there are four points in plane #1, each indistinguishable from each other). This sum (3), then, is .
The additional lengths created when the plane is added; which in turn is comprised of:
(2) The sum of the lengths of the lines in the plane (the same as (1)); PLUS
(3) The sum of the distances of each of the points in plane #1 from each of the points in plane #2.
Thus, when we add the totals (1), (2) and (3) together, we get:
If I understand you correctly, I think you are mistaken here.
Now, when I place another cube right smack alongside the first cube, there are now 12 planets connected by lines of force.
The sum of these lines, if I'm not mistaken, is:
When we add four more 'planets' in plane #3, we get, in addition to the distances above:
(4) The sum of the lengths of the lines in plane #4; this is, of course, the same as (1) and (2); namely . PLUSSo I reckon that the total for the 12 planets is:To see the pattern being created here, break the expression above into three components:
The sum of the all the distances of from each of the points ; which comprises:
(5) the distances of the points in plane #2, which is the same as (3) above; namely ; and
(6) the distances of the points in plane #1, which is (taking the distances from and multiplying by 4) .
(1) for plane #1and notice that (1), (2) and (4) are the same; (3) and (5) are the same; (6) is the new expression formed from the distances of the points in the first plane from the last. So we can re-arrange these as:i.e.So when the next new plane is added comprising points , there will be additional totals equal to:
(2) + (3) for plane #2
(4) + (5) + (6) for plane #3
(4) + (5) + (6) for the points in planes #2, #3 and #4; PLUSSo this, when re-arranged, is equivalent to:... and so on. I hope you can see the pattern emerging. I realise that I'm still quite a way short of developing a formula for the total with planes, but at least you should be able to see how to continue - if you still want to, that is!
(7) the distances from plane #1 to plane #4; namely .
No, I'm sure that the sum won't be the same if the cubes are not arranged in a straight line.
AND: In this scheme of things, the cubes line up in a straight line. Would the sum of lengths be the same if the cubes were arranged differently (but always so that the xth cube is added in such a way that four of its corners line up exactly with the four corners of one of the other cubes)?