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• Level: GCSE
• Subject: Maths
• Word count: 1874

# Black and white squares

Extracts from this document...

Introduction

## By: M.T.Tawil

Part 1: Investigate to see how many squares would be needed to make any cross shape made in this way.

## Sequence Number

### Total of squares

Difference between the  White squares added

## 4

1 BK: 0 W     Total  = 1

1 BK: 4 W     Total   = 5

5 BK: 8 W      Total    = 13

13 BK: 12 W   Total   = 25

Theory used in counting the squares

25 BK: 16 W   Total  = 41

41 BK: 20 W   Total   = 61        Predicting next three sequences using the theory

61 BK: 24W   Total  = 85

#### Sequence 8

85 BK: 28W   Total   = 113

#### Sequence 9

113 BK: 32W   Total   = 145

#### Sequence 10

145 BK: 36 W   Total   = 181

I have proved my predictions with the diagram’s above.

Finding formulas for the squares

#### The white squares

Add on the consistent difference between the white squares, which in this case is four.

Formula = 4+N

I am going to test this formula with an example.

Sequence number = 56

Consistent difference = 4

The number of white squares = 4+56= 60

The black squares

I would have take in to thought, that the black squares of a pattern equal the total number of squares in the previous pattern, this helps in the search of a formula.

We also know that the difference between the black squares and the total of all the squares is 4.

So therefore we could say that to find the black squares, we should takeaway the formula of the white squares from the total squares.

Middle

20

60

41

86

7

85

231

24

84

61

147

1

Tn-2nd

Wn-2

Bn-2

2

Tn-1st

Wn-1

Bn-1

3

Tn

Wn

Bn

After producing the formulas above, I have decided to look at the issue in a different angle, by adding the cumulative total number of squares, cumulative number of white and cumulative number of black, I have also decided to place the result of my founding’s in a formula form with the values being represented by algebraic letters, if my theories prove right I will prove them by substituting the letters with true values.

The letters in the table represent the algebraic letters I spoke of earlier, now the first obvious matter I have noticed is the plain fact that the total of white and black in a certain pattern will give us the total number of squares in that pattern, e.g.   in pattern 6, the total of white squares which is 20 + the total of black squares which is 41 will gives us the total of squares in that pattern which is 61, even though this is plainly simple I have made a formula relating to this.

### Tn-2  = Wn-2 + Bn-2

My second observation, concerns the fact which I found in my table, that the sum we get from taking away the number of white squares in certain pattern from the total squares of that same pattern, we see that this sum is equal to the sum found when we add the white and black squares from the previous pattern. Io have created a formula to show this fact and have the evidence below it.

### Tn – 1  - Wn – 1  = Wn – 2 + Bn – 2

E.g.

Pattern6    61-20 = Pattern5   25+16

Pattern5   41-16 = Pattern4   12+13

My third observation, concerns finding the number of white squares in a pattern, which is found when we take away the total number of squares of a pattern from the total number of squares of the pattern prior to it.

## Tn – 1 – Tn – 2 = Wn – 1

E.g. (Pattern 6) 61- (Pattern 5) 41= 20 white squares in pattern 6

My third observation, concerns finding the number of black squares in a pattern, which is simply the total of black squares in pattern equals to the total of squares in the pattern prior to it.

## Bn  = Tn – 1

E.g. (Black squares in pattern 6) 41 = 41 (Total number of squares in pattern 5)

Now, I will come to the main point, which is to find a formula for the total number of squares with out any information apart from the pattern number, and through this formula I will be able to find out how many squares would be needed to make any cross shape made in the patterns above.

##### Formula

n(n-1)          x 4 + 1

2

Proving the formula

n = 1

1x(1-1)= 0

0/2= 0

0x4= 0

0+1= 1

n = 2

2x(2-1)= 2

2/2= 1

1x4= 4

4+1= 5

n = 3

3x(3-1)= 6

6/2= 3

3x4= 12

12+1= 13

n = 4

4x(4-1)= 12

12/2= 6

6x4= 24

24+1= 25

n = 5

5x(5-1)= 20

20/2= 10

10x4= 40

40+1= 41

n = 6

6x(6-1)= 30

30/2= 15

15x4= 60

60+1= 61

n = 7

7x(7-1)= 42

42/2= 21

21x4= 84

84+1= 85

I believe now that I have proven my initial objective, and it is possible to find the total number of squares for any cross section shape like the ones above.

Part 2: Extend your investigatons to three dimensions.

## Sequence Number

### Total of cubes

Difference between the White cubes added

## 8

Conclusion

Now I have two equations with only (a) and (b). Now I will cancel (b) to obtain the value of (a) using algebra.

14 = 15a + 3b                 2 = 3a+ b

-  6 =   9a+ 3b             x 3

-  8 =   6a                           6 = 9a+ 3b

8 = 6a

4 / 3 = 3a / 3

a = 1⅓

Now with the value of (a), I can find the value of (b) using substitution.

2 = 3 x 1⅓ + b

2 = b + 4

2 - 4 = b + 4 - 4

b = -2

Having obtained a, b and d, the value of c can be solved using algebra.

n = 1, T = 1

T = an³ + bn² + cn + d

1 =  1⅓ x 1³+ -2 x 1² + c x 1 + -1

1 = 1⅓ x 1+ -2 x 1 + c - 1

1 = 1⅓+ -2 + c - 1

1 = 1⅓ - 2 + c - 1

1 = c + 1⅓ - 3

1 = c - 1⅔

1 + 1⅔ = c - 1⅔ + 1⅔

c = 2⅔

I now have the general term for this sequence.

T = 1⅓n³ - 2n² + 2⅔n - 1

T = 1⅓n³ - 2n² + 2⅔n - 1

Proving the formula

n = 1

1⅓ x 1³ - 2 x 1² + 2⅔ x 1 - 1 = 1

1⅓ x 1 - 2 x 1 + 2⅔ - 1 = 1

1⅓ - 2 + 1⅔ = 1

n = 2

1⅓ x 2³ - 2 x 2² + 2⅔ x 2 - 1 = 7

1⅓ x 8 - 2 x 4 + 5⅓ - 1 = 7

10⅔ - 8 + 4⅓ = 7

n = 3

1⅓ x 3³ - 2 x 3² + 2⅔ x 3 - 1 = 25

⅓ x 27 - 2 x 9 + 8 - 1 = 25

36 - 18 + 7 = 25

n = 4

1⅓ x 4³ - 2 x 4² + 2⅔ x 4 - 1 = 63

1⅓ x 64 - 2 x 16 + 10⅔ - 1 = 63

85⅓ - 32 + 9⅔ = 63

n = 5

1⅓ x 5³ - 2 x 5² + 2⅔ x 5 - 1 = 129

1⅓ x 125 - 2 x 25 + 13⅓ - 1 = 129

166⅔ - 50 + 12⅓ = 129

n = 6

1⅓ x 6³ - 2 x 6² + 2⅔ x 6 - 1 = 231

1⅓ x 216- 2 x 36 + 16 - 1 = 231

288 - 72 + 15 = 231

n = 7

1⅓ x 7³ - 2 x 7² + 2⅔ x 7 - 1 = 377

1⅓ x 343- 2 x 49+ 18⅔  - 1 = 377

457⅓- 98+ 17⅔   = 377

I believe now that I have proven my initial objective, and it is possible to find the total number of cubes for any 3D cross section shape like the ones above.

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