4 4
This shows there is a relation, and the formula produced will be a quadratic in the form:
Un = an² + bn + c
U1 = a + b + c = 5
U2 = 4a + 2b + c = 13
U3 = 9a + 3b + c = 25
U4 = 16a + 4b + c = 41
1) U – U1 = 3a + b = 8
2) U3 – U1 = 8a + 2b = 20
3) U4 – U1 = 15a + 3b = 36
4) U4 = 5a + b = 12
3
4) – 1) = 2a= 4 => a = 2
Putting a = 2 into 1) = 6 + b = 8 => b = 2
If a = 2 and b = 2, U1 = 2 + 2 + c = 5 => c = 1
Giving the formula Un = 2n² + 2n + 1
Checking this formula, using a cross I have already drawn, one of width 7:
7/2 + 0.5 = 4
(2 x 16) + (2 x 4) + 1
= 32 + 8 + 1
= 41
which is the total found by drawing out the cross-shape, proving the formula.
From evidence in the table, the formula for the number for the number of white squares can be found as it is just an arithmetic progression, the number increasing by 4 each time. This gives the formula to find the number of white surrounding squares as:
Un= 4n
Following on from this, the total number of squares, with the total of white surrounding squares removed, gives the total of shaded squares for each width.
Un = 2n² + 2n + 1 – 4n
= 2n² - 2n + 1
And this result is the same as the previous total number of squares, the size and shape of the cross increasing regularly:
5/2 + 0.5 = 3 3/2 + 0.5 = 2
18 – 6 +1 = 13 8 + 4 + 1 = 13
From this I can find out if my formula for the total number of squares is correct by looking at two different ways of getting the next term in the series:
2(n + 1)²+ 2(n + 1) + 1 must = 2n² + 2n + 1 + 4(n + 1)
LHS = 2(n² + 2n + 1) + 2n + 2 + 1
=2n² + 4n + 2 + 2n + 2 + 1
= 2n² + 6n + 5
RHS = 2n² + 2n + 1 + 4n + 4
=2n² + 6n + 5
And this proves that the formula is correct and works for all values of n.
Extending this investigation into 3 dimensions means I will again have to look at the totals of squares in the same way as for 2 dimensions. The 3 dimensional shape is built up in layers, with the width, n, being the middle layer and this decreasing above and below to 1 square e.g.:
Layer 1
Layer 2
Layer 3
This shows how simply the 3 dimensional shape is made up. When the size increases, layer 2 will appear again, and a larger layer will be the middle layer. This gives a table of results as:
This idea of layers can be expressed as:
n-1
Un =2 ( E (2i² + 2i. + 1)) + 2n² + 2n + 1
1
Meaning that as another bigger layer is added, it’s total number of squares, and another total number of squares for the previous width, to keep the 3 dimensional diamond shape.
Another way of approaching this is to look at the differences between the totals again. For the total number of squares:
1 7 25 63 129
6 18 38 66
12 20 28
- 8
This shows there is a cubic relation between the totals, which can be found by doing similar simultaneous equations to those in the 2 dimensional investigation:
Un = an³ + bn² + cn + d
U1 = a + b + c + d = 7
U2 = 8a + 4b + 2c + d = 25
U3 = 27a + 9b + 3c + d = 63
U4 = 64a + 16b + 4c + d = 129
U2 – U1 = 7a + 3b + c = 18 = U5 U5 x2 = 14a + 6b + 2c = 36 = U8
U3 – U1 = 26a + 8b + 2c = 56 = U6 U5 x3 = 21a + 9b + 3c = 54 = U9
U4 – U1 = 63a + 15b + 3c = 122 = U7
U6 – U8 = 12a + 2b = 20 = U10 U10 x3 = 36a + 6b = 60 = U12
U7 – U9 = 42a + 6b = 68 = U11
U11 – U12 = 6a = 8
(4/3 x 12) + 2b = 20
- 16 + 2b = 20
- 2b = 4
- b = 2
(4/3 x 7) + (2 x 3) + c = 18
4/3 + 2 + 8/3 + d = 7
This gives the formula:
Un = 4n³+ 6n²+ 8n + 3
3
To prove this formula, I need to find the formula for finding the number of white squares so I can find the next term in the series two ways, and therefore prove the formula works:
6 18 38 66
12 20 28
- 8
This shows there is a squared relation between these totals, and the simultaneous equations are used again.
Un = an² + bn + c
U1 = a + b + c = 6
U2 = 4a + 2b + c = 18
U3 = 9a + 3b + c = 38
U2 – U1 = 3a + b = 12 = U4 U4 x2 = 6a + 2b = 24 = U6
U+3– U1 = 8a + 2b = 32 = U5
U5 – U6 = 2a = 8
(3 x 4) + b = 12
4 + 0 + c = 6
Giving the formula:
U7 = 4n² + 6
From this a proof of the formula for Un can be found:
4(n+1)³+ 6(n+1)²+ 8(n+1) + 3 must = 4n³+6n²+ 8n + 3 + 12(n+1)²+ 18
3 3 3
LHS = 4(n+1)(n+1)²+ 6(n²+2n+1) + 8n +11
3
= 4(n+1)(n²+2n+1) + 6n²+12n+6 + 8n + 11
3
= 4(n3+2n2+n+n2+2n+1) + 6n2 + 12n + 6 +8n +11
3
= 4n3 + 8n2 + 4n + 4n2 + 8n + 4 + 6n2 + 12n + 6 + 8n +11
3
= 4n3 + 18n2 + 32n + 21
3
RHS = 4n³+6n²+ 8n + 3 + 12(n+1)²+ 18
3
= 4n3 + 6n2 + 8n + 3 + 12 (n2+2n+1) + 18
3
= 4n3 + 6n2 + 8n + 3 + 12n2 + 24n + 18
3
= 4n3 + 18n2 + 32n + 21
3
