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• 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

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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

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