I can see that the 2nd difference has the same number which means I do not to go on and finding any more differences as there isn’t because it’s constant as there are two differences I will have to use quadratic sequence as well as linear sequence to solve the equation. To find the quadratic sequence I will use the equation below
tn = an² + b n + c
Where it says tn, this equals the total number of squares.
In the formula above first I will work out a n², c and then b n
To find a I have to take my 2nd difference which is 4 and then divide it by 2 which is 4/2= 2, so therefore a in the formula is replaced by 2, the formula should now look like:
tn = 2n² + b n + c
To find c I have to find the zeroth term, to do this I will put my results in different so it makes it easier to get the zeroth term and I shall explain how I went about finding it.
n: 1 2 3 4 5 6
1 1 5 13 25 41 61
+0 +4 +8 +12 +16 +20
+4 +4 +4 +4
As you can see above there is pattern as I also explained earlier that I can see to get the next no of squares the first no of squares is added to the 1st difference so therefore I must go back and find out the first square came, the illustration above shows how I found the zeroth term. I found that if you 1 to 0 you will get one for the next sequence or I could have also went backward from the second difference e.g. 12, 8 , 4, 0 then find the no of squares.
The zeroth term has been solved and its 1 so you take this number and replace with c in the formula, our formula now should look like:
tn = 2n² + b n + 1
Now to complete the formula we have to find the last one which is b in the equation:
Below is the working out of how to find b, I will first do the working out for the 2nd term which is 5 and then the other terms will be easy to do when the formula has been formed
tn = 2x1x1 + b n + 1 = 5 (1x1 = 1 then 2x1=2)
I did this using BODMAS which starts with brackets if there are no brackets then you go on to the next one which is powers or indices etc…
tn = 2 + b n + 1 = 5
Then you use BODMAS again, because there are no brackets, powers or multiply in the equation above we have to use addition which is:
2+1= 3 then I did
3-5 = -2 which is b in the equation
So therefore the equation is complete and will now look like:
tn = 2n² - 2n + 1
Testing the rule
As you can see above I have now got a complete formula which I will check to see if it works
Sequence two
-
2(22) -4 +1
- 2(4) -4 +1
- 8 – 4 +1
= 5 Correct
As you can see the formula is working I am going to try it on some more sequences to be 100% sure that it is fully working and has no errors
Sequence 3
-
2(32) -6 +1
- 2(9) -6 +1
- 18 – 6 +1
= 13 Correct
Sequence 6
-
2(62) -12 +1
- 2(36) -12 +1
- 72 -12 +1
= 61 Correct
The formula that I found seems to be working fine without any errors as proven above. I am now going to use this formula to find the number of squares (tn) in a higher sequence:
Sequence 11
-
2(112) -22 +1
- 2(121) -22 +1
- 242-22 +1
= 221 Correct
To check if I am correct or incorrect I will start from sequence 6 as the 1st difference for every pattern is 4 and the 2nd difference is constant. I can see that +4 is added to each pattern in the 1st difference so to do this I will add the total number of squares with the 1st difference for sequence 6 through its previous number and then I will add the 2nd difference which is contact +4 the same applies to all other sequences, the table below should explain everything
In the table above the 4 in the middle column is the constant difference which is difference two which always stays the same for every sequence that I am doing and the one where it says 20,24,28,32,36 which goes up in 4’s so four is added to its previous number and then the total number of squares which 61, 85 113, 145, 181 these are taken from the previous total and added on to make the next sequence.
By doing all of these checking’s, the sequence 11 appears to be correct as shown above and therefore the formula is correct.
A quicker way of finding the formula for the 2D sequence can also be this way
General formula an² + b n + c
When n=1 a+b+c = 1 (1)
When n=2 4a+2b+c =5 (2)
When n=3 9a+3b+c =13 (3)
(2)-(1) 3a+b =4 (4)
(3)-(2) 5a+b =8 (5)
(5)-4) 2a =4
÷2 ÷2
a =2
Finding b
Substitute a=2 into equation 5
5a+b=8
5x2+b=8
10+b=8
-10 from both sides therefore b = - 2
Finding c
Substitute a=2 and b=-2 into equation 3
9a+3b+c=13
9x2+3x-2+c=13
18-6+c=13
12+c=13
-12 from both sides to leave c on its own and that will help find c, therefore c = +1
Complete formula: 2n² -2n +1
Finding the formula for the black squares (border)
n 1 2 3 4 5 6
0 4 8 12 16 20
+4 +4 +4 +4 +4
There is no 2nd different for this because the 1st difference is constant because it’s the same as +4 added to the no of black squares.
To find the formula I will have to use the term number and the difference, multiply the two numbers and then takeaway 4 as the difference is the same from forwards and backwards.
Therefore the formula will be 4n-4 (LINEAR SEQUENCE)
I will use this formula to find some terms in the table to see if it’s working
E.g. to find the 2nd term I will have to take the term no which is 2 and then multiply by the difference which is 4 and then -4 so 4*2= 8 -4 = term 2 = 4 correct
n term 5
5*4= 20 – 4 = 16 black squares Correct
n term 6
6*4= 24 – 4 = 20 black squares Correct
Prediction
From the formula 4n-4 I am going to predict that sequence 10 will have 36 black squares. To see if my prediction was correct I am going to check it using the formula 4n-4
10*4= 40 – 4 = 36 Prediction was Correct
I am now going to use the formula 4n-4 to find a higher sequence
n term 15
15*4= 60 – 4 = 56 black squares Correct
n term 100
100*4= 400 – 4= 396 black squares Correct
As you can see I have the formula 4n-4 is working fine without any errors.
Conclusion
I have found out that the 2D shapes pattern was quadratic this is shown by the 2nd difference 4 which is constant and is a necessary element of any quadratic sequence. Because the constant difference was 4 I knew that it had to include 2 in it as you have to divide it by two. There was also linear sequence involved as I was doing the formula for the border, which was far simpler than the quadratic. As I was doing my investigation I had to find a pattern between the total number of squares, where I then had to create a formula, make predictions and test it to see if it was working. It turned out that the formula was working fine which can be used to get any sequence without having to draw the shape, I also noticed that the total number of squares and white squares were odd numbers therefore I knew that if it had even number in it, it would mean that there is something wrong but for the black squares (border) it was the opposite, as they were even numbers which meant that if it had odd number it will be wrong.
Extending my Investigation to 3 Dimensional
Now that the 2D investigation is completed and the formula is working I am now going to continue my investigation to 3D shapes and try to find the formula for it
Drawing the 3D Shapes
Shape 1= 1 white square, 0 black squares
Shape 2= 1 white square, 6 black squares
Shape 3= 7 white squares, 18 black squares
Shape 4= 25 white squares, 38 black squares
Shape 5= 63 white squares, 66 black squares
The table shows the number of white cubes and the number of black cubes to make it clear of what I am doing
Finding the Formula for the Total no of Cubes
To find out which formula to use I am going to do the same as I did to the 2D shapes, I am going to find the differences between the numbers as shown below, also from the table above I get a pattern sequence that looks like the one shown below
n 1 2 3 4 5 6
1 7 25 63 129 231
+6 +18 +38 +66 +102
+12 +20 +28 +36
+8 +8 +8
In the table above the differences do not become constant until the third row which means the formula is cubic (n3). We use the same as the previous one for 2D but in 3D it’s more complicated.
General formula
nth term = an3 + bn2 + cn + d
because there are a+b+c+d in the formula above to find therefore we have to use 4 terms as shown below and its called simultaneous equation.
When n = 1 a(1)3+b(1)2+c(1)+d =1
a + b + c + d =1 (1)
When n = 2 a(2)3+b(2)2+c(2)+d =7
8a + 4b + 2c + d =7 (2)
When n = 3 a(3)3+b(3)2+c(3)+d =25
27a + 9b + 3c + d = 25 (3)
When n = 4 a(4)3+b(4)2+c(4)+d =63
64a + 16b + 4c + d = 63 (4)
Simultaneous Equations
For the first I am going to show the full working out, so it’s easier to understand
(2)-(1) 8a + 4b + 2c + d=7
- a + b + c + d=1
= 7a+3b+c =6 (5)
(3)-(2) 19a+5b+c =18 (6)
(6)-(5) 12a+2b =12 (7)
(4)-(3) 37a+7b+c =38 (8)
(8)-(6) 18a+2b =20 (9)
(9)-(7) 6a =8
6a =8
÷6 ÷6
We can’t do it this way as it comes in decimals therefore we have to do it in fractions
8÷2
6÷2 = 4/3 therefore a= 4/3
Substituting a= 4/3 into equation (9) to find b
Which is
18a+2b=20
18x4/3+2b= 20
=72/3+2b= 20
=24+2b=20
24-24+2b=20-24
2b= -4
To leave b on it’s on I have to divide both sides by two to and then that will help me to find b
So that’s
2b/2= b -4/2= -2 therefore b= -2
To find c substitute a = 4/3 and b = -2 into equation (8)
37a+7b+c=38
37x4/3+7(-2)+c=38
=148/3-14+c=38
148 14
3 1
X3
=
148 42 = 106/3
3 3
106/3 + c = 38
106/3 - 106/3 + c = 38 - 106/3
38-(106÷3)= 8/3 so therefore c = 8/3
To find d substitute a = 4/3, b = -2 and c = 8/3 into equation (1)
a+b+c+d=1
4/3-2+8/3+d=1
4-2+d=1
2+d=1
2-2+d=1-2 which is -1 therfore d = -1
Now I have my complete formula= 4/3n3 - 2n2 + 8/3n – 1
Formula 4/3n3 - 2n2 + 8/3n – 1 checking
n 2 = 7
4/3(2)3 – 2(2)2 + 8/3(2) – 1=7
4/3(8) – 2(4) + 16/3 – 1 = 7
32/3 -8 + 16/3 -1 = 7
32/3+16/3= 16
16-8= 8-1= 7 CORRECT
n 5 = 129
4/3(5)3 – 2(5)2 + 8/3(5) – 1=129
4/3(125) – 2(25) + 40/3 – 1 = 129
500/3 -50 + 40/3 -1 = 129
500/3+40/3= 180
180-50= 130-1= 129 CORRECT
My formula is correct, and therefore it should work on all the sequences as proven above.
