### The Borders

The shape on the left is a dark cross shape that has been surrounded by white squares to create a bigger cross shape.

The shape consists of 25 small squares in total. To make the next shape in the sequence, white squares are added to the existing shape, creating a bigger cross shape made up in the same way. For my investigation, I am going to work out how many squares would be needed to make up any cross shape built up in this way by working out a general formula. Then I am going to extend my investigation to 3 dimensions.I intend to use structure to work out a general formula.2 Dimensional InvestigationI am going to start by building a 1×1 cross-shape and adding on the borders. The complete shape will be known as the cross-shape (this includes the border squares).

I have decided to define this shape as a 1×1 shape.1×1 shape = 1 square2x2 shape = 5 squares3x3 shape = 13 squares4x4 shape = 25 squares5x5 shape = 41 squares6x6 shape = 61 squaresI now have enough data to analyse my results:Number in sequence (n)Total number of squares (tn)1125313425541661To work out which formula to use I will now put my results in a table showing the differences between the numbers:n tn 1st difference 2nd difference1 142 5 483 13 4124 25 4165 41 4206 61I 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 = anï¿½ + 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(ï¿½ 2) a = 2(substitute into (4)) 6 + b = 4(- 6) b = -2(substitute into (3)) 18 – 6 + c = 1312 + c =13(- 12) c = 1Therefore my formula is: tn = 2nï¿½ – 2n + 1Independent Check:When n = 4, tn should equal 252(4ï¿½) – 2(4) +132 – 8 + 1 =25My 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)2n(n+1) = 1(1+1) = 12 2n(n+1) = 2(2+1) = 32 2n(n+1) = 3(3+1) = 62 2As 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 2n(n -1) = 1(1-1) = 02 2n(n – 1) = 2(2 -1) = 12 2n(n – 1) = 3(3 -1) = 32 2n(n – 1) = 4(4 -1) = 62 2The 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)ï¿½ + 2(n – 1) + 1 =2(nï¿½ – 2n + 1) + 2n – 2 =2nï¿½ – 4n + 2 + 2n – 2 + 1 =2nï¿½ – 2n + 1The 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 is2nï¿½ – 2n + 1 .3 Dimensional InvestigationNow 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 3×3 cross (13 cubes), two 2×2 crosses (2×5 cubes), and two 1×1 crosses (2×1 cubes).

There are 25 cubes in all.A 4x4x4 shape would be made up of one 4×4 cross (25 cubes), two 3×3 crosses (2×13 cubes), two 2×2 crosses (2×5 cubes), and two 1x1crosses (2×1 cubes). There are 63 cubes in all.If I add up all the cubes for every 3D cross shape I get:Number in sequence (n)Total number of cubes (tn)112732546351296231To 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 difference1 162 7 1218 83 25 2038 84 63 2866 85 129 361026 231The 3rd difference is the same, so therefore tn = anï¿½ + bnï¿½ + 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(ï¿½6) a =8/6 = 1((substitute into (9)) 24 + 2b = 20(-24) 2b = -4(ï¿½2) b = -2(substitute into (5)) 9( – 6 + c = 63( + c = 6(-5() c = 2((substitute into (4)) 85( – 32 + 10( + d = 6364 + d = 63(-56) d = -1Therefore my formula is: tn = 1(nï¿½ -2nï¿½ + 2(n + 1Independent Check:When n = 5, tn should equal 1291( (5ï¿½) – 2(5ï¿½) + 2((5) + 7166( – 50 + 13( – 1 = 129My formula is correct for this sequence.Mel Spencer 10LMaths Coursework