16
5 41 4
20
6 61
I can clearly see that the 2nd difference is the same, which means that my sequence is a quadratic sequence. The formula to find a quadratic sequence is tn = an² + bn + c, where tn equals the total number of squares.
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) a = 2
(substitute into (4)) 6 + b = 4
(- 6) b = -2
(substitute into (3)) 18 - 6 + c = 13
12 + c =13
(- 12) c = 1
Therefore my formula is: tn = 2n² - 2n + 1
Independent Check:
When n = 4, tn should equal 25
2(4²) - 2(4) +1
32 - 8 + 1 =25
My formula is correct for this sequence, but to make sure my formula is accurate, I am now going to prove it using structure.
When I look at any of the cross shapes, I see that they can be split up into four triangles, with one square left in the middle:
The formula for the sequence of triangular numbers (1, 3, 6, 10 etc.) is n(n+1)
2
n(n+1) = 1(1+1) = 1
2 2
n(n+1) = 2(2+1) = 3
2 2
n(n+1) = 3(3+1) = 6
2 2
As there are 3 squares in each of the triangles, and 3 is the 2nd number in the triangular numbers sequence, I know that if I use the standard formula for finding the triangular numbers, it will not work, because the cross shape is 3rd in the sequence, not 2nd. This means that I have to use a different formula.
If instead of using n(n+1) I use n(n -1) ,this is what I find:
2 2
n(n -1) = 1(1-1) = 0
2 2
n(n - 1) = 2(2 -1) = 1
2 2
n(n - 1) = 3(3 -1) = 3
2 2
n(n - 1) = 4(4 -1) = 6
2 2
The numbers in this sequence are the triangular numbers, but instead of 3 being 2nd in the sequence, it is now 3rd. This means that I can use my new formula, because both numbers I am putting into the equation are 3rd in their sequence.
2(n - 1)² + 2(n - 1) + 1 =
2(n² - 2n + 1) + 2n - 2 =
2n² - 4n + 2 + 2n - 2 + 1 =
2n² - 2n + 1
The formula I have found using structure is the same as the formula I found using quadratics. Therefore I conclude that the formula needed to work out how many squares would be needed to make up any cross shape is
2n² - 2n + 1 .
3 Dimensional Investigation
Now I have found the formula for the 2D shape, I will move on to find the formula for the 3D shape. A 3x3x3 3D cross shape is made up like this:
The 3x3x3 shape is made up of one 3x3 cross (13 cubes), two 2x2 crosses (2x5 cubes), and two 1x1 crosses (2x1 cubes). There are 25 cubes in all.
A 4x4x4 shape would be made up of one 4x4 cross (25 cubes), two 3x3 crosses (2x13 cubes), two 2x2 crosses (2x5 cubes), and two 1x1crosses (2x1 cubes). There are 63 cubes in all.
If I add up all the cubes for every 3D cross shape I get:
To work out which formula to use I will again put the numbers in a table, showing the differences:
n tn 1st difference 2nd difference 3rd difference
1 1
6
2 7 12
18 8
3 25 20
38 8
4 63 28
66 8
5 129 36
102
6 231
The 3rd difference is the same, so therefore tn = an³ + bn² + cn + d.
When n = 1 a + b + c + d = 1 (1)
When n = 2 8a + 4b + 2c + d = 7 (2)
When n = 3 27a + 9b + 3c + d = 25 (3)
When n = 4 64a + 16b + 4c + d = 63 (4)
(2)-(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
(÷6) a =8/6 = 1
(substitute into (9)) 24 + 2b = 20
(-24) 2b = -4
(÷2) b = -2
(substitute into (5)) 9 - 6 + c = 6
3 + c = 6
(-5) c = 2
(substitute into (4)) 85 - 32 + 10 + d = 63
64 + d = 63
(-56) d = -1
Therefore my formula is: tn = 1n³ -2n² + 2n + 1
Independent Check:
When n = 5, tn should equal 129
1 (5³) - 2(5²) + 2(5) + 7
166 - 50 + 13 - 1 = 129
My formula is correct for this sequence.