• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6
7. 7
7
8. 8
8
9. 9
9
10. 10
10
11. 11
11
12. 12
12
13. 13
13
14. 14
14
• Level: GCSE
• Subject: Maths
• Word count: 1244

# Borders Coursework

Extracts from this document...

Introduction

Candidate no. 1874                                                                                       Saagar Kotecha 11SZ

Borders Coursework

Part 1:

Aim:

I shall investigate a set pattern of squares and shall look at how the total number of squares increases each time the geometric shape gets larger. I shall then look at the relationship between the increase in size of the geometric shape and the number of additional squares that need to be added to cause this increase.

 Pattern Number Shape Squares 1 Black Squares   = 1White Squares   = 0Total Squares    = 1 2 Black Squares   = 1White Squares   = 4Total Squares    = 5 3 Black Squares   = 5White Squares   = 8Total Squares    = 13 4 Black Squares   = 13White Squares   = 12Total Squares    = 27 5 Black Squares   = 25White Squares   = 16Total Squares    = 55 6 Black Squares   = 41White Squares   = 20Total Squares    = 61

## Prediction:

As the cross increases in length and width, I predict more squares will have to be added. I predict that the number of white squares should be directly proportional to the size of the cross.

Middle

Formula:                 + bn + c

a = 2nd difference

2

a =   4/2

=  2

.

.    .       =     2n² + bn + c

Formula: 2n² +                 + c

1                 5               13               25                41                61

2×(1²)                   2×(2²)                   2×(3²)                       2×(4²)                    2×(5²)                 2×(6²)

= 2                     = 8                           = 18                       = 32                    = 50                         = 72

1-2                     5-8                         13-18                         25-32                     41-50                 61-72

= -1                    = -3                = -5                     = -7                    = -9                 = -11

-2                        -2                    -2                           -2                           -2

.

.    .       =     2n² – 2n + c

Formula: 2n² – 2n +

When n = 1

2 × (1²) – (2 × 1)

= 2 – 2

= 0 + ? = 1

= 0 + 1 = 1

c = 1

.

.    .  Formula for the total number of squares =  2n² – 2n + 1

Check:

formula for black squares + formula for white squares = formula for total no. of squares

2n² – 6n + 5       +             4n – 4                =              2n² – 2n + 1

=                 2n² – 2n + 1                        =             2n² – 2n + 1

Test:

 7 Black Squares   = 61White Squares   = 24Total Squares    = 85

When n= 7

Number of Black Squares = 2n² – 6n + 5

= 2(7²) – 9(7) + 5

= 98 – 42 +5

= 56 + 5

= 61

Number of Black Squares = 4n – 4

= 4(7) – 4

= 28 – 4

= 24

Total Number of Squares  = 2n² – 2n + 1

= 2(7²) – 2(7) + 1

= 98 – 14 + 1

= 85

Geometric Proof:

After examining the shapes it was realised that it is possible to work out the total number of squares geometrically. If the total number of white squares on one side are counted and the total number of

Conclusion

;height:108.8px;margin-left:0px;margin-top:0px;" alt="image01.png" />

Total Cubes    = 7

3

Total Cubes    = 25

4

Total Cubes    = 63

5

Total Cubes    = 129

6

Total Cubes    = 231

Finding the n thterm for the total number of cubes:

Formula:  an³ + bn² + cn + d

1                7               25                63              129              231

6                    18               38                66               102

12                        28                     36                   44

8                           8                          8

Formula:                 + bn² + cn + d

a = 3rd difference

6

a =   8/6

=  4/3

.

.    .       =     4/3n³ + bn² + cn + d

Formula: 4/3n³ + bn² + cn + d

1                 7               25               63                129              231

4/3×(1³)                4/3×(2³)                  4/3×(3³)                       4/3×(4³)                    4/3×(5³)                 4/3×(6³)

= 4/3                     = 32/3                   = 36                       = 256/3                    = 500/3                 =288

1-4/3                     7-32/3                 25-36                         63-256/3                     129-500/3                 231-288

= -1/3            = -11/3                = -11                     = -67/3                    = -113/3                 = -57

-1/3         -11/3          -11             -67/3           -113/3             -57

10/3           22/3            34/3             46/3             58/3

4                       4                  4                    4

Formula:  4/3n³ +                  + cn + d

a = 2nd difference

2

a = 4/2

= 2

.

.    .       =    4/3n³ + 2n² + cn + d

Formula: 4/3n³ + 2n² + cn + d

-1/3          -11/3           -11            -67/3           -113/3             -57

2×(1²)                   2×(2²)                   2×(3²)                       2×(4²)                    2×(5²)                2×(6²)

= 2                     = 8                           = 18                       = 32                    = 50                        = 72

-1/3+2                   -11/3+8                  -11+18                   -67/3+32                   -113/3+50                 -57+72

= 5/3                  = 13/3                 = 7                   = 29/3                   = 37/3                 = 15

8/3                        8/3                    8/3                           8/3                           8/3

.

.    .       =     4/3n³ – 2n² + 8/3n + d

Formula: 4/3n³ + 2n² + 8/3n +

When n = 1

[4/3 × (1³)] + [2 × (1²)] + [8/3 × 1]

= 4/3 + 2 + 8/3

= 6

= 6 + ? = 7

= 6 + 1 = 7

d = 1

.

.    .  Formula for the total number of cubes = 4/3n³ + 2n² + 8/3n + 1

Check:

When n = 3

4/3n³ + 2n² + 8/3n + 1

4/3(3³) + 2(3²) + 8/3(3) + 1 = 63

4/3(27) + 2(9) + 8 + 1 = 63

36 + 18 + 8 + 1 = 63

63             = 63

.

.    . Formula is correct

This student written piece of work is one of many that can be found in our GCSE T-Total section.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related GCSE T-Total essays

1. ## T-total coursework

5 star(s)

an upside-down L shape), the formula will not work, as one of the terms is missing and the total (5n+5x+5wy-7w) is not reached. Part 3 Use grids of different sizes again. Try other transformations and combinations of transformations. Investigate the relationships between the T-total, the T-number, the grid size, and the transformations.

2. ## T-Shapes Coursework

I will express the T-Number as 'n'. n-19 n-18 n-17 n-9 n 2 3 4 12 21 1 2 3 11 20 n-19 n-18 n-17 n-9 n We can see a pattern that applies to both T-Shapes, found by using the nth term, and so we are going to use these T-shapes to help us find the formula.

1. ## T-Shapes Coursework

I believe this t-total will be 323 - 5 = 318. 51 + 60 + 68 + 69 + 70 = 318. I am sure that the rule of moving any rotated t-shape forward or backwards you just increase or decrease the t-total by 5 or more and it effectively works.

2. ## T-Shapes Coursework

43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92

1. ## Maths Coursework- Borders

Pattern no. No. of black squares No. of white squares Total no. of squares 1 1 0 1 2 1 4 5 3 5 8 13 4 13 12 25 5 25 16 41 6 41 20 61 7 61 24 85 From this I predict that the next pattern (pattern no.8)

2. ## Borders - Investigation into how many squares in total, grey and white inclusive, would ...

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 8 This shows there is a cubic relation between the totals, which can be found by

1. ## T-Total Coursework

The seven works with all sizes of grids. I will demonstrate now that the formula 5*N-63 is correct: The T-Total of 23 is 52. To check rightness of my formula I will use it: 5*23 - 63 = 115 - 63 = 52 To be completely sure I am going to use the formula to find out some other results.

2. ## Maths Coursework T-Totals

We know with will not make a difference to the final answer as proved in question 2. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33

• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to