Total
If you look at stage 2 you are able to see that the square is split into two different sections diagonally. The black line marks the section with squares with 3 squares in it and the red line marks the section with squares with 2 squares in it. That means the square is made up of three 3x3 sections and two 2x2 sections.
I then went to look at the next stage in the sequence. I divided it up into sections the same way as the previous shape. This time I was able to divide it into 4 sections made up of 4 squares and 3 sections made up of 3 squares. I noticed that in the smaller section, the number of sections and the number of squares in that section were equal to the stage (N) and that the number of sections and squares of the larger sections were the sequence number +1. Using each of the sections I made 2 squares;
If I find the areas of these squares and add them to one another I get; (N+1)(N+1)+(N)(N)=2N²+2N+1
Shaded
In stage 2, if you take the shaded squares from the sequence and paired them off into sets of 2, you are left with 4 sets of 2 squares.
Then in the 3rd stage in the sequence, if you pair off the shaded squares into sets of 3 you are left with 4 sets of 3 squares. This tells us that if you multiply the stage number (N) by 4, you are given the amount of shaded squares in the shape. Therefore; Nth term for Shaded = 4N
Unshaded
To find the amount of unshaded squares all I have to do is subtract the shaded squares from the total amount of squares;
2N²+2N+1-4N = 2N²-2N+1
Conclusion
I can now conclude that my formulae are definitely correct as I have made predictions and found them to be true and also spatially justified my formula to find out why they are true.
I will now investigate how a formation of shaded and unshaded hexagons grows. I have chosen hexagons as, like squares, they tessellate, meaning that when put together, the corners meet to form 360°. I will now draw the first four stages of this growth to establish if any patterns arise.
Finding the Nth Term
I am now going to put the information I found from each of the shapes into tabular form so that I can understand it better and perhaps spot a pattern emerging. If I find a pattern I will use it to determine an Nth term, meaning a term that will be applicable to all shapes of this nature.
Perimeter
I can see a clear pattern here, the sequence for perimeter is going up regularly in 12’s, and therefore this sequence is a linear sequence and has an Nth term.
Nth term = 12N+?
From my table I can see that I need to add on 4 each time to get the number in the sequence so;
Nth term for Perimeter = 12N+6
Shaded
I can see a clear pattern here, the sequence for shaded is going up regularly in 6’s, and therefore this sequence is a linear sequence and has an Nth term.
Nth term = 6N+?
From my table I can see that 6 times N gives me the number in the sequence so; Nth term for Shaded = 6N
Unshaded
I can see here that there is not must of a pattern in the 1st difference, but when I calculate the 2nd difference I can see that it goes up in 6’s, therefore this is a quadratic sequence and has an Nth term. The second difference is 6 therefore the coefficient of N² must be half of 6 i.e. 3.
Nth term = 3N²+bN+c
From my table I will solve the Nth term using simultaneous equations, I will use stages 2&3 to solve the Nth term for the unshaded squares.
1) 12+2b+c=7
2) 27+3b+c=19
2-1 is 15+b=12
b=-3
To find c: 12-6+c=7
c=1
Nth term for Unshaded = 3n²-3n+1
Total
Total is equal to Shaded plus Unshaded so;
6N
+ 3n²-3n+1
3n²+3n+1
Nth term for Total = 3n²+3n+1
Predictions
The formulae I have found are:
Perimeter 12N+6
Shaded Squares 6N
Unshaded Squares 3n²-3n+1
Total Squares 3n²+3n+1
I will now do a check to see if my formulae are correct. I will test this by predicting, using my formulae, the perimeter, number of shaded squares, number of unshaded squares and the total number of squares, for the next two shapes in this sequence, stages 5 and 6. I will then calculate each of these manually and compare the answers.
Stage 5
I predict:
Perimeter - 12(5)+6=66
Shaded - 6(5)=30
Unshaded - 3(5)²-3(5)+1=61
Total - 30+61=91
Stage 6
I predict:
Perimeter - 12(6)+6=78
Shaded - 6(6)=36
Unshaded - 3(6)²-3(6)+1=91
Total - 36+91=127
I found that my predictions for stages 5 and 6 were correct, as I have counted the hexagons and found the answers to be the same as what I had predicted.
Spatial Justification
Now that I have found that my formulae are correct by making predictions then proving them, I will now use diagrams to spatially justify my formula and explain why these formulae work.
Perimeter
In stage 1 if we count the sides we get 18, but if we separate the shapes up into 4 different sections we find that there are two sets of 6 and two sets of 3, which can be made into one group of 12 and 6 extra shapes.
If we do the same with stage 2, we see that there are two sets of 12 and 2 sets of 3, which can be made into 2 groups of 12 and six extra shapes. We can see here that the stage number (N) multiplied by 12 plus the six extra sides give us the perimeter. Therefore; Nth term for perimeter = 12N+6
Shaded
If you take stage 1 you will notice that there are 6 shaded hexagons which make up one group of 6.
In stage 2 you can see that there are 12 shaded hexagons, making up two groups of 6. So we can clearly see that the stage number (N) multiplied by 6 give you the amount of shaded shapes in the formation. Therefore; Nth term for shaded = 6N
Total
If you take stage 1 and divide it into columns as I have here, we find that there are two columns of 2 and one of 3 which means that:
2x2=4
1x3=3
4+3
=3+3+1
=3x1²+3x1+1
=3N²+3N+1
2 3 2 3 3
4 4
5
When you do the same for stage 2, we find that there are two columns of 3, two columns of 4 and one column of 5. which means that:
2x3=6
2x4=8
1x5=5
6+8+5
=6+12+1
=3x2²+3x2+1
=3N²+3N+1
Therefore this works in all cases and therefore; Nth term for total = 3N²+3N+1
Unshaded
To find the amount of unshaded hexagons all I have to do is subtract the shaded hexagons from the total amount of hexagons;
3n²+3n+1-3n²-3n+1=6N
Conclusion
I can now conclude that my formulae are definitely correct as I have made predictions and found them to be true and also spatially justified my formula to find out why they are true.
So far I have found two sets of formulae, I will now use these formulae to try and determine formulae that will work with any growing shape that tessellates.
I will now try to prove my theory by predicting the shaded, unshaded and total for triangles using these general formulae.
My predictions are:
Shaded 3n
Unshaded 1.5n²-1.5n+1
Total 1.5n²+1.5n+1
Finding the Nth Term
I am now going to put the information I found from each of the shapes into tabular form so that I can understand it better and perhaps spot a pattern emerging. If I find a pattern I will use it to determine an Nth term and see if my predictions are right.
Shaded
I can see a clear pattern here, the sequence for shaded is going up regularly in 3’s, and therefore this sequence is a linear sequence and has an Nth term.
Nth term = 3N+?
From my table I can see that 3 times N gives me the number in the sequence so; Nth term for Shaded = 3N
Unshaded
I can see here that there is not much of a pattern in the 1st difference, but when I calculate the 2nd difference I can see that it goes up in 3’s, therefore this is a quadratic sequence and has an Nth term. The second difference is 3 therefore the coefficient of N² must be half of 3 i.e. 1.5.
Nth term = 1.5N²+bN+c
From my table I will solve the Nth term using simultaneous equations, I will use stages 2&3 to solve the Nth term for the unshaded squares.
1) 6+2b+c=4
2) 13.5+3b+c=10
2-1 is 7.5+b=6
b=-1.5
To find c: 6-3+c=4
c=1
Nth term for Unshaded = 1.5N²-1.5N+1
Total
Total is equal to Shaded plus Unshaded so;
3N
+ 1.5N²-1.5N+1
1.5N²+1.5N+1
Nth term for Total = 1.5N²+1.5N+1
I have now found that my general formulae are correct as I used them to predict the shaded, unshaded and total for triangles and have found that my predictions are correct. Now that I know that my general formulae are correct I have no need to continue to investigate 2D shapes that tessellate, as I can use my general formulae on any of them.
I have decided, now that I can investigate 2D shapes no further, I will investigate 3D shapes. I will choose cubes because it is the 3D version of squares and will give me answers that will be easy to work with, therefore making it the logical choice. Each cube grows by the adding another cube to each of its exposed faces.
I have noticed that each shape is made up off sections, these sections are similar to the shapes that I got when I was investigating squares at the very start of the growing shapes investigation. For example stage 1 in the cube sequence is made up off two “start shapes” and stage 1 in the square sequence. Shape 2 in the cube sequence is made up off two “start shapes”, two stage one’s and stage 2 from the square sequence. I believe this pattern will continue.
Using this information I will be able to get the total number of cubes for the first five stages of this sequence using the information from the squares tables. Now I am going to try and find a formula for total number of cubes in each shape.
Total Cubes
I can see here that there is not must of a pattern in the 1st difference or the 2nd difference, but when I calculate the 3rd difference, I can see that it goes up in 8’s, therefore this is a cubic sequence and has an Nth term. The third difference is 8 therefore the coefficient of N must be a sixth of 8 i.e. 4/3
Nth term = 4/3N+bN²+cN+d
From my table I will solve the Nth term using simultaneous equations with three equations and three unknowns to find co-efficient’s b, c and d. First I will find my three equations:
N=1 4/3(1)+b(1)²+c(1)+d=7 x3 to remove the fractions
4+3b+3c+3d=21
N=2 4/3(2)+b(2)²+c(2)+d=25 x3 to remove the fractions
32+12b+6c+3d=75
N=3 4/3(3)+b(3)²+c(3)+d=63 x3 to remove the fractions
108+27b+9c+3d=189
Now that I have found my 3 equations I will use them to find b, c, and d using simultaneous equations and therefore obtain the rule for the Nth term.
Equation A - 4+3b+3c+3d=21
Equation B - 32+12b+6c+3d=75
Equation C - 108+27b+9c+3d=189
B-A=D
32+12b+6c+3d=75
- 4+3b+3c+3d=21
28+9b+3c=54
So equation D is: 9b+3c=26
C-B=E
108+27b+9c+3d=189
- 32+12b+6c+3d=75
76+15b+3c=114
So equation E is: 15b+3c=38
I will now subtract equation D from E to give me the value of b which I then can use to find c and d.
15b+3c=38
-9b+3c=26
6b=12
b=2
I will now find c by substituting it into equation D.
9b+3c=26
18+3c=26
3c=8
c=8/3
This now allows me to find d by substituting this information into equation A.
4+3b+3c+3d=21
3(2)+3(8/3)+3d=17
6+8+3d=17
3d=3
d=1
So Nth term for volume of cubes is = 4/3N+2N²+8/3N+1
I will now test this by checking that, when I have N=5 I will get the answer 231.
4/3(5)+2(5)²+8/3(5)+1
166 2/3+50+13 1/3+1=231
Prediction
I will now do a check to see if my formula is correct. I will test this by predicting, using my formula, the volume of the next shape in this sequence, stage 6. I will then calculate this manually and compare the answers.
I predict:
4/3(6)+2(6)²+8/3(6)+1
288+72+16+1=377
Following the link that I discovered between squares and cubes earlier, I can tell that, stage 6 should be made up of two start shapes, two stage 1’s, two stage 2’s, two stage 3’s, two stage 4’s, two stage 5’s and a stage 6.
And if we add all of the squares in each of these shapes together, we get:
1+1+5+5+13+13+25+25+41+41+61+61+85=377
This means that my prediction is correct! Therefore my formula is correct!
Conclusion
At the start of this investigation I was given 3 shapes, the start shape, stage one and stage two of the squares sequence and was told to investigate. The first thing I did was to draw the first six stages of the sequence and calculate the perimeter, shaded, Unshaded and total amount of squares in each stage. I then put this into a table to see if any patterns would emerge. When patterns did emerge I used them to find the Nth term. I then checked my formulae by predicting the next two stages of the sequence and proving my predictions. After that I spatially justified my formula to prove why they were correct.
Next I did the same thing for hexagons and found another set of formulae, I then compared the formulae for squares and the formulae for hexagons to see if there were any connections between them. I found a connection and used this to determine the general formulae for any 2D shape that tessellates.
Next I decided to see if my general formulae worked, I decided to use triangles, as, like squares and hexagons they tessellate. I checked my formulae by predicting the shaded, unshaded and total amount of triangles for the first three stages of the sequence. My predictions were correct and my general formula worked. I now believed that I had investigated 2D shapes as much as I could, and could now use my general formula for any 2D shape, I then decided to move on to 3D shapes. I chose the cube as it is very similar to the square and would be the simplest 3D shape to investigate. Now I investigated the volume of the cubes, I discovered that the cubes were made up of the square sequences built on top of each other. I used this information to create a formula for the volume of the cubes. I then used my formula to predict the next stage in the cube sequence. And found my prediction to be correct. Here is a summary of what I found:
Squares
Perimeter 8N+4
Shaded Squares 4N
Unshaded Squares 2N²-2N+1
Total Squares 2N²+2N+1
Hexagons
Perimeter 12N+6
Shaded 6N
Unshaded 3n²-3n+1
Total 3n²+3n+1
Triangles
Shaded 3n
Unshaded 1.5n²-1.5n+1
Total 1.5n²+1.5n+1
General formulae for 2D
Perimeter (2x)N+x
Shaded xN
Unshaded (½x)N²-(½x)N+1
Total (½x)N²+(½x)N+1
Cubes
Volume 4/3N+2N²+8/3N+1
