One of the first things I noticed by just looking at these results were that the differences were all perfect squares x 10. This would probably prove to be of some significance in future.
I had also noticed that the pattern started with only the second term as a one by one square only consisted of one square, a minimum four being needed. My aim now was to find a rule for this sequence.
10 40 90 160
30 50 70 1st Difference
20 20 2nd Difference
As this was proving to be a quadratic sequence with two differences, I divided the 2nd difference by 2! to determine the number that was to be the coefficient of n2.
so
I then took 10n2 off each term so I had a linear sequence left
So shown on a table I had:
Formula:
-20 -20 -20
Formula for the linear part:
So put together, the complete formula would be:
(Simplifying down to this)
At this point, without any further calculation, I noticed two things:
The beginning part of the expression, 10, also happened to be the length of the grid. I hypothesized these two were directly connected, that this number would be whatever the grid length was.
I also realized that if instead of writing the expression down as if written as, one of the s may well be the length, while the other is the width if working with rectangles instead of squares. Until now I have been working with only the difference in products of corners in squares, i.e. 2x2, 3x3, etc. so as the length of a square is equal to its width, the two could both be represented using a single expression.
I tested out my first assumption by redoing the process on an 11x11 grid instead of 10x10.
2x2:
(2x12) - (1x13) = 11
(5x15) – (4x16) = 11
3x3:
(36x56) – (34x58) = 44
(40x60) – (38x62) = 44
4x4:
(81x111) – (78x114) = 99
(86x116) – (83x119) = 99
The pattern came out as:
So,
11 44 99 176
33 55 77 1st Difference
22 22 2nd Difference
, determining the coefficient to be used for n2.
Formula:
-22 -22 -22 Formula for linear part:
Put together:
(Simplifying down to this)
Indeed the only part of the end formula that changed, was the part hypothesized to be directly linked to the length of the grid. With the grid length 10, the formula came out to be, with the grid length as 11, it came out to be. As this part of the formula appeared to be linked to the grid length, it could now be replaced with an expression representing the grid length so it would work for any grid. Thus the formula now became. If the assumption that one of the s were in fact the length, and the other the width continued, the formula then became.
To prove this formula, I looked at the situation algebraically and came up with a list of expressions to represent the real life values.
The grid, its numbers, and the numbers encompassed within the rectangle of a certain size, could be looked at in this way:
If is thought of as, the other values,, and, if expressed with the variables given, and, in relation to , they come out to be:
Now all there is to do with these expressions is the same thing we would do to them if they were numbers, find the products of the pairs of corners, then find their difference, in other words, .
When the respective expressions are substituted for,, and , it comes out to:
This eventually does in fact simplify down to, as predicted.
To be absolutely sure, I tested the formula.
The example I used for testing:
On a grid of 8 length, the corner numbers in a 5x3 rectangle.
(14x26) – (10x30) = 64
Using the formula:
Both answers come out the same, meaning the formula worked.
To extend the investigation further, I started thinking what would happen on a 3D grid, with cubes and cuboids instead of squares and rectangles.
By a 3D grid I meant one like so:
where it is a line of grids of some dimension. In this drawing of course, I have the grids spaced out to make easier to understand.
A cube/cuboid I’d be working with would look something like this on the grid:
And the corners in question would be these:
One set of corners would be,, and.
The other would be,, and.
This time I went straight to expressing the points on the diagram algebraically as it seemed a lot quicker. The variables used were as follows:
If , then the other values came out as:
The calculation concerning the first set of corners was -. When the respective expressions were substituted for these values, it came out to:
Simplifying down to
This was my first 3D formula.
Now for the second set, the calculation was -. When the respective expressions were substituted for these values, it came out to:
Simplifying down to
This was my second 3D formula.
I tested the both of these to check that everything was ok with them.
The example I came up for the test was:
In the example, , , , , , , and . Other than that, , , , , .
For the first formula,
If done with the formula instead,
For the second formula,
If done with the formula instead,
In both cases, the answer came out the same, both formulas worked. One slight problem was that somewhere in the process I managed to mix up the order I subtracted in so that my first formula always gave a negative answer (as I subtracted a larger value from a smaller one) and my second formula always gave a positive answer (as I subtracted a smaller value from a larger one, like I was supposed to). This isn’t that much of a problem as what the actual number was would always came out the same, the sign in front of it solely depends on the order in which the numbers were subtracted from one another. As the investigation asks for a mere difference, the sign could be ignored altogether.
