Position in sequence: 0 1 2 3 4 5 6
No. Of Squares ( c ): 1 1 5 13 25 41 61 . . . . . (I)
First Differences ( a + b ): 0 4 8 12 16 20 . . . . . . (ii)
Second Differences (2a): 4 4 4 4 4 . . . . . . . . (iii)
The bottom row of differences indicates a constant number, which shows there to be a pattern. If the fourth row had not indicated a constant number pattern (i.e. 2,2,2,2 or 6,6,6,6) then I would keep increasing the rows until I found one. If I was struggling to find a constant number, I would gather more information. This is what happened with my 3D investigation.
If n is the position in the sequence, 2a = 4 (iii) (therefore) a = 4 / 2 a = 2.
a + b = 0 (ii) and a = 2
We have 2 + b = 0 b = -2 and c = 1 (I).
Substituting the above into the formula a n 2 + b n + c, we have:
a n 2 + b n + c
2 n 2 - (2n) + 1
2 n 2 - 2n + 1
Here we have a formula, 2 n 2 - 2n + 1, for calculating the nth term in the sequence.
Testing The Formula
I can test the formula by using it on previous shapes, which I have built out of the Lego bricks. I will test five of my results. Below are my testing methods and my results.
So far, all of the Cross-shapes that I have tested, have been correct. This shows that the formula works for these shapes. I can’t, however, prove that it will work for any shape, though by testing some different sized cross-shapes, I can discover how many total squares they consist of and then draw them out, to see if the formula was correct. This will still not prove that it works for any number, however, by extending my two dimensional investigation, I will whether is more reliable. This will go further to proving that the investigation works with any sized shape, though will not make it certain.
Below are two other cross shapes. They are supposed to be 85 and 113 squares. I will now draw them out to test that this is correct.
An 11x11 Cross-shape
a n 2 + b n + c
2 x 7 2 - 2 x 7 + 1
Answer: 85
Counted Answer: 85
Status: Both Correct
A 13x13 Cross-shape
a n 2 + b n + c
2 x 8 2 - 2 x 8 + 1
Formula Answer: 113
Counted Answer: 113
Status: Both Correct
Summary
There is no way to make sure that the formula will work for every sized shape but I tried to make it more reliable by showing that it works with all the ones tested and two other cross-shapes that I had not investigated before, which were all correct.
The only way to be certain that the formula works with every sized cross shape is by testing it on every possible sized one, which would be very difficult. This is why I feel that I have generated enough information to show that the formula that I have found out, 2 n 2 - 2 n + 1, gives the expected results.
( 3D Investigation on next page )Three Dimensional Investigation
Introduction
I will now investigate to find the three dimensional formula. I am going to start by building a 1x1 cross-shape and adding on the borders. The complete shape will be known as the cross-shape (this includes the border squares). I again used Lego bricks to make the cross-shapes. The larger diagrams for the cross-shapes in 3D are too complex to see all the squares on. The cross-shape below is 1x1 with border squares around it and you can see all but the middle square.
Obtaining Results
I obtained my results by building this three dimensional cross-shape up as in the two dimensional investigation. The three-dimensional cross-shapes are more complex than the two-dimensional cross-shapes, due to the 3D ones having another dimensional to draw.
I used Lego bricks again, with two different colours for the border squares and the darker squares. These helped me to gather my data. This is because they give a visualisation of the problem. This is easier to understand than trying to work it out in my head.
Analysing Results
I think that investigating five different cross-shapes will give me enough information to find a formula. This is because the information I gathered will give me enough data to analyse and utilise my results. The data I have found out is shown in the table below. I have done one less shape size in this investigation because the 3D shapes required more bricks than I had. So I had to stop at a 7x7 shape size.
From these results I cannot see a direct relationship with the amount of border squares. I will investigate this further as I think that this relationship will be important in finding the formula to discover how many squares are required to surround the darker cross-shape thus making a bigger cross shape.
One relationship with the results is that the original cross-shapes total squares is the same number as the total number of squares on the previous shape. For example, the total squares for a 5 x 5 cross-shape squares (including border squares) is 63 and the original Cross-shape squares (not including the border) for a 7 x 7 Cross-shape is also 63.
Also the total amount of squares is always odd and the amount of border squares is always even with the darker squares always totaling up to be an odd number.
Finding a Formula
I will apply the difference table method to attempt to find a formula for this investigation. The results are shown below:
Position in sequence: 0 1 2 3 4 5
No. Of Squares: -1 1 7 25 63 129 . . . . . . . (I)
First Differences: 2 6 18 38 66 . . . . . . . . (ii)
Second Differences: 4 12 20 36 . . . . . . . . (iii)
Third Differences: 8 8 8 . . . . . . . . . . . . . (iiii)
The bottom row of differences indicates a constant number, which shows there to be a pattern. If the fourth row had not indicated a constant number pattern (i.e. 2,2,2,2 or 6,6,6,6) then I would keep adding rows until I found one. If I were struggling to find a constant number, I would gather more information. This is what happened with my 3D investigation. Due to the complexity of the structure, the first time I did the experiment I only did three tests. 1x1, 3x3 and 5x5. This was not enough information for me to discover a direct link, so when I got to several rows down, I decided to gather more information.
We can now use the equation a n 3 + b n 2 + cn + d to find the formula. So now we have to find the values for the variables so we can put them into the equation.
We already know the value of n which indicating the position of the cross-shape in the sequence, so we can find out what a, b, c and d’s values are.
But from this I am unable to determine a formula, consequently I will look at another method.
I have noticed that the shape is made up from a series of 2D shapes. This is shown in the diagram below:
Fig-1
This would indicate that I could use some sort of combination of my 2D formula (2n2+bn+c).
This is the standard formula: (2 n2 – 2 n + 1) + 2
This formula works out the area for the 2D shape (the middle one in the above diagram) and then adds 2 to the answer because every three dimensional shape (apart from the single square) has two end pieces. This makes the correct answer for the shape shown below.
As you can see, two end pieces (the +2 of the formula) are added on to either end to make the shape on the right hand side. This formula works for this shape only (as it has only the 2D layer and the two end pieces).
But this formula will not work for a shape with more layers. So to find the formula for more layers, I have worked out that this extra information has to be added to the equation to take account of the additional layers.
The extended formula is shown below:
(2 n2 – 2 n + 1) + 2 x [(2(n-1) 2 -2(n-1)+1)] + 2
This formula will only work with the 3rd term in the sequence. Below is the formula broken down to show why the mathematics is there:
This works out the 2D shape: (2 n2 – 2 n + 1)
This works out the Second Layer: 2 x [(2(n-1) 2 -2(n-1)+1)]
This is the value of the End Squares: 2
So when they are joined together they form the formula :
(2 n2 – 2 n + 1) + 2 x [(2(n-1) 2 -2(n-1)+1)] + 2
so we can test this formula by doing the calculations for the above diagram.
n = 3
2D (centre) Layer + 2x Second Layers + 2x End Squares
= (2 n2 – 2 n + 1) + 2 x [(2(n-1) 2 -2(n-1)+1)] +2
= (2 x 32 – 2 x 3 + 1) + 2 x [(2x22 –2x2)+1)] +2
= (2 x 9 – 6 + 1) + 2 x [(8 – 4)+1)] +2
= (18 – 6 + 1) + 2 x (4+1) +2
= (12 + 1) + 2 x 5 +2
= 13 + 10 +2
= 13 + 10 + 2
= 25
This also works out as predicted
To check further I will calculate one more pattern, when n=5. The diagram of this is shown below.
This time there will be another term in the calculation to calculate the additional layer. The formula for this will be:
(2 n2 – 2 n + 1) + 2 x [(2(n-1) 2 -2(n-1)+1)] + 2 x [(2(n-2) 2 -2(n-2)+1)] + 2
Below is the working out for the above diagram. There is another layer this time because the 3D shape’s size has increased. So my prediction for this shape is that it will have a total square amount of 63.
2D (centre) Layer + 2x Second Layers + 2x Third Layers + 2x Ends
= (2 n2 – 2 n + 1) + 2 x [(2(n-1) 2 -2(n-1)+1)] + 2 x [(2(n-2) 2 -2(n-2)+1)] +2
= (2 x 42 – 2 x 4 + 1) + 2 x [(2x32 –2x3)+1)] + 2 x [(2(4-2) 2 -2(4-2)+1)] +2
= (2 x 16 – 8 + 1) + 2 x [(18 – 6)+1)] + 2 x [(2x4 –2x2+1)] +2
= (32 – 8 + 1) + 2 x (12+1) + 2 x (8 -4+1) +2
= 24 + 1 + 2 x 13 + 2 x 5 +2
= 25 + 26 + 10 +2
= 25 + 26 + 10 + 2
= 63
Both the count of the squares and the formula’s answer were correct. So from all of my investigation into this formula I now believe that I can predict what the total amount of squares for shape n will be.
Below is an example of how to find the 5th term. I have calculated the answer as 129, because I built it with the Lego bricks. I will now attempt to calculate it with my formula. Firstly, I will apply the formula 2n-1 to discover how many layers there are. 2 x 5 - 1 = 9 which shows that there are 9 layers. I can take three layers off this because I know that there will be one 2D layer, the middle one and two end squares. So I will need to work out 6 layers values. Below is the formula:
(2 n2 – 2 n + 1) + 2 x [(2(n-1) 2 -2(n-1)+1)] + 2 x [(2(n-2) 2 -2(n-2)+1)] + 2 x [(2(n-3) 2 -2(n-3)+1)] + 2
I will now work this formula out:
(2 n2 – 2 n + 1) + 2 x [(2(n-1) 2 -2(n-1)+1)] + 2 x [(2(n-2) 2 -2(n-2)+1)] + 2 x [(2(n-3) 2 -2(n-3)+1)] + 2
= (2 x 52 – 2 x 5 + 1) + 2 x [(2(5-1) 2 -2(5-1)+1)] + 2 x [(2(5-2) 2 -2(5-2)+1)] + 2 x [(2(5-3) 2 -2(5-3)+1)] + 2
= (2 x 25 – 10 + 1) + 2 x [(2(4) 2 -2(4)+1)] + 2 x [(2(3) 2 -2(3)+1)] + 2 x [(2(2) 2 -2(2)+1)] + 2
= (50 – 10 + 1) + 2 x [(2x16 –2x4+1)] + 2 x [(2x9 –2x3+1)] + 2 x [(2x 4 –2x2+1)] + 2
= (40 + 1) + 2 x (32 –8+1) + 2 x (18 –6+1) + 2 x (8 –4+1) + 2
= 41 + 2 x (24+1) + 2 x (12+1)] + 2 x (4+1) + 2
= 41 + (2 x 25) + (2 x 13) + (2 x 5) + 2
= 41 + 50 + 26 + 10 + 2
= 129
Again this agrees with my calculation and I now believe that this formula could calculate all the possible sequences.
I have also observed that there may be a method of calculating how many layers there are in the sequence number
The above diagram shows shape 2, which is a 3D, 1x1 cross shape built up of three layers. This shape is number 2 in the sequence. It has 3 layers and the formula 2n-1, with the value of n being the term in the sequence (in this case 2) so the formula ends up as 2(2)-1 which is 3. This is correct in this case, as there are 3 layers and the formula is the same value.
For the third sequence, I will use the same formula, 2n-1. This time n will be the value of 3. So the formula will become 2(3)-1 which equals 5. The diagram over shows what the value is.
As the diagram shows, there are five layers. The formula 2n-1 is so far successful. I will try one more just to check the formula. I will try a 5x5 cross-shape, there should be 2n-1 layers. So by substituting the value of n for 4 (the term in the sequence is 4), the formula now reads 2(4)-1, which equals 7. So I predict that there will be 7 layers to the 5x5 cross-shape. The diagram for this shape is shown below.
This is also give the expected answer.
Summary
Due to there being too many variables in the equation a n 3 + b n 2 + cn + d I was unable to resolve the problem in this manor. Instead I decided to use a pattern I noticed while investigating the Lego brick models. I think the new method, which I have developed, could be simplified with further development to give a much more concise formula.
There is no way to make sure that the formula will work for every sized shape but I tried to make it more reliable by showing that it works with the ones I built with Lego bricks and one other cross-shapes that I had not investigated before, which were all correct.
The only way to be certain that the formula works with every sized cross shape is by testing it on every possible sized one, which would be very difficult. This is why I feel that I have generated enough information to show that the formula that I have found out gives the expected results.
