How many squares on a chess board

This is my first piece of mathematics coursework. In this project I will be finding out how many squares on a 8×8 chess board. In this project I will be including labelled drawings, clear enough to allow a non mathematician to follow, clearly written work, initial drawings of each of my steps and a finished formula justified. Finally an extension on my board making it rectangular.

Plan

* Introduction

* Spider diagram

* I will be collecting data from different sized grids ranging from 2×2 to 8×8.

* Explanation of how to add up the squares on the grid

* Further calculations

* Create a table of values.

* Formula and formula justified

* Draw a graph

* Extension

* Conclusion

* Evaluation

Tree diagram

Explanation of how to add up the squares on the grid

1. Count how many singular squares there are

=16

2. Count how many double squares there are.

=9

3. Continue this until you complete the whole shape in this case 4×4.

4. Once you have completed this you should have 208 squares on the 8×8 grid.

Further calculations

1×1=1

2×2=5

3×3=14

4×4=30

5×5=………

1×1

1×1=1 =1

2×2

2×2=4

(2-1)x(2-1)=1×1=1 =4+1=5

3×3

3×3=9

(3-1) (3-1)=2×2=4

(2-1) (2-1)=1×1=1 =9+4+1=14

4×4

4×4=16

(4-1)x(4-1)=3×3=9

(3-1)x(3-1)=2×2=4 =16+9+4+1=30

(2-1)x(2-1)=1×1=1

Check 4×4

2 2 2 2

= (4) + (4-1) + (3-1) + (2-1)

= 16+9+4+1

=30

Use of Algebra

2 2 2 2

(nxn) = (n) + (n-1) + (n-2) + (n-3)

Estimating 5×5

2 2 2 2 2

= (4) + (5-1) + (5-2) + (5-3) + (5-4)

= 25+16+9+4+1

= 55

Table of values

x y D1 D2 D3

1 1

+4

2 5 +5

+9 +2

3 14 +7

+16 +2

4 30 +9

+25 +2

5 55 +11

+36 +2

6 91 +13

+53 +2

7 140 +15

+62

8 204

Finding the formula

The cubic formula of type

3 2

Y= Ax + Bx + Cx + D

(x+1)

3 2

Y= Ax1 + Bx1 + Cx1 + D=1

A+B+C+D=4 1

(x+2)

3 2

Y= Ax2 + Bx2 + Cx2 + D=5

8A+4B+2C+D=5 2

(x+3)

3 2

Y= Ax3 + Bx3 + Cx3 + D=14

27A+9B+3C+D =14 3

(x+4)

3 2

Y= Ax4 + Bx4 + Cx4 + D=30

64A+16B+4C=D=30 4

The four equations to be used are:

1. A+B+C+D=1

2. 8A+4B+2C+D=5

3. 27A+9B+3C+D=14

4. 64A+16B+4C=D=30

4-3= 37A+7B+C=16 5

3-2= 19A+5B+C=9 6

2-1= 7A+3B+C=4 7

5-6= 18A+2B=7 8

6-7= 12A+2B=5 9

8-9= 6A=2=0.33’=1

3

Sub A= 1 in 9

3

1 A+2B=5

3

(12×1)+2B=5

3

4+2B=5

2B=5-4

2B=1

B= 1

2

Sub A and D in 7

7A+3B+C=4

(7) (1) + (3) (1) + C=4

3 2

7+3+C=4

3 2

C= 4-7-3

1 3 2

C= 4×6 – 7×2 – 3×3

1 3 2

C= 24 – 14 – 9

6 6 6

C= 24-14-9

6

C= 1

6

Sub A,B,C in 1

A+B+C+D=1

1 + 1 + 1 + D=1

3 2 6

D= 1 – 1 – 1 – 1

3 2 6

1×6 – 1×2 – 1×3 – 1

1 3 2 6

6 – 2 – 3 – 1

6 6 6 6

6-2-3-1

6

=0

Justifying the formula

X=4

3 2

Ax + Bx + Cx + D

3 2 3 2

(1) (4) + (1) (4) + (1) (4) + 0 OR 4 + 4 + 4

3 2 6 3 2 6

64 + 16 + 4 = 2

3 2 6 3

64×2 + 16×3 + 2×2

3 2 3

128 + 48 + 4 + 0

6 6 6

128 + 48 + 4 + 0

6

= 180 = 30

6

Justifying the formula of a higher number

3 2

20 + 20 + 20

3 2 6

=2870 Squares

Number of squares in a rectangle

3×1

3×1=3

(3-1)x(1-1)=0 =3

3×2

3×2=6

(3-1)x(2-1)= 2×1=2

(2-1)x(1-1)=0 =8

3×3

3×3=9

(3-1)x(3-1)= 2×2=4

(2-1)x(1-1)= 1×1=1 =14

3×4

3×4=12

(3-1)x(4-1)= 3×2=6

(2-1)x(3-1)= 1×2=2

(1-1)(2-1) =21

The sequence number in this arrangement:

Stage 1 Stage 2 Stage 3

Rectangle: 3×2 3×3 3×4

Sequence: 8 14 20

+6 +6

Stage 1 =8

Stage 1= 1×6=6+2 =8

Formula is (x6+2)

N th rule= nx6+2

=6n+2

The reason in this that you don’t get square numbers as answers is because the shapes are not square.

Extension 2

The cubic formula of type

3 2

Y= Ax + Bx + Cx + D

1. A+B+C+D=2

2. 8A+4B+2C+D=8

3. 27A+9B+4C+D=20

4. 64A+16B+4C+D=40

4-3= 37A+7B+C=20 5

3-2= 19A+5B+C=12 6

2-1= 7A+3B+C=6 7

5-6= 18A+2B=8 8

6-7= 12A+2B=6 9

8-9= 6A=2

2/6= 0.33’=1

3

Sub A= 1 in 8

3

18A+2B=8

18Ax 1 +2B=8

3

6+2B=8

2B= 8-6

2B=2

B= 1

Sub A ; B in 7

7A+3B+C=6

7x 1 + 1×3+C=6

3

2 1 A+ 3B+C=6

3

6-5= 2

3

C=2

3

Sub A, B ; c in 1

A+B+C+D=2

1 + 1 +2 + D =2

3 3

D=0

Conclusion

The results of my investigation has lead me to believe the following conclusion. As the size of the grid 2×2 increases to 8×8, so does the number of squares. Using my algebraic equation the formula was obtained. When tested against a known number it seemed to be working satisfactory.

Evaluation

At the beginning of my project the work seemed to be very simple but as the project progressed I found it got harder and harder. If I hade more time to do my project I would have continued on my extension and maybe, tried to work out how many squares in a triangle, and a formula for working that out. In this project I have gained much more knowledge about numbers, shapes and creating formulas.