(2-1)x(2-1)=1x1=1
Check 4x4
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 5x5
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
(12x1)+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= 4x6 – 7x2 – 3x3
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
1x6 – 1x2 – 1x3 - 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
64x2 + 16x3 + 2x2
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
3x1
3x1=3
(3-1)x(1-1)=0 =3
3x2
3x2=6
(3-1)x(2-1)= 2x1=2
(2-1)x(1-1)=0 =8
3x3
3x3=9
(3-1)x(3-1)= 2x2=4
(2-1)x(1-1)= 1x1=1 =14
3x4
3x4=12
(3-1)x(4-1)= 3x2=6
(2-1)x(3-1)= 1x2=2
(1-1)(2-1) =21
The sequence number in this arrangement:
Stage 1 Stage 2 Stage 3
Rectangle: 3x2 3x3 3x4
Sequence: 8 14 20
+6 +6
Stage 1 =8
Stage 1= 1x6=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
- A+B+C+D=2
- 8A+4B+2C+D=8
- 27A+9B+4C+D=20
- 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 + 1x3+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 2x2 increases to 8x8, 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.
