4 + 2B + C = 49
so 2B + C = 45
we now solve the simultaneous equation
B + C = 63
2 B + C = 45
∴- B = 18 OR B = - 18
if B = - 18 then C = 63 – (-18)=81
so the formula is
yn = n – 18n + 81
if we factories the equation yn = (n-9) (n-9) ∴yn = (n-9)
To prove the formula is correct use any term in the sequence, I will use the fourth term.
Y4 = 4 – 18 x 4 +81
=16 – 18 x 4 +81
= 25
It can be seen from these results that the fourth term on and 8x8 board is indeed 25thus proving the formula is correct.
Using the factorized formula yn (n-9) we can further prove using the fifth term
Yn = (5-9) = 4 = 16 it can be seen from the table the fifth term is 16.
The second investigation involves counting the number of non-over lapping rectangles of a given size on a certain size chessboard.
RESULTS
2x2 board
1x2=2
3x3 board
1x2=4
1x3=3
2x3=1
4x4 board
1x2=8
1x3=5
1x4=4
2x3=2
2x4=2
3x4=1
5x5 board
1x2=12
1x3=7
1x4=6
1x5=5
2x3=3
2x4=2
2x5=2
3x4=1
3x5=1
4x5=1
6x6 board
1x2=18
1x3=12
1x4=8
1x5=7
1x6=6
2x3=6
2x4=4
2x5=3
2x6=3
3x4=2
3x5=2
3x6=2
4x5=1
4x6=1
5x6=1
7x7 board
1x2=24
1x3=16
1x4=10
1x5=9
1x6=8
1x7=7
2x3=6
2x4=4
2x5=4
2x6=3
2x7=3
3x4=3
3x5=2
3x6=2
3x7=2
4x5=1
4x6=1
4x7=1
5x6=1
5x7=1
6x7=1
8x8 board
1x2=32
1x3=20
1x4=16
1x5=11
1x6=10
1x7=9
1x8=8
2x3=10
2x4=8
2x5=5
3x6=5
2x7=4
2x8=4
3x4=4
3x5=3
3x6=2
3x7=2
3x8=2
4x5=2
4x6=2
4x7=2
4x8=2
5x6=1
5x7=1
5x8=1
6x7=1
6x8=1
7x8=1
A linear sequence is found using the 1x5 size rectangle the sequence is 5,7,9,11
To find the formula for this sequence I used
Yn = dn+B
Where d=the common difference, n is the term to be found and B is the first term of the sequence minus the common difference, i.e 5-2 = 3
Yn = 2n + 3
Because we have started on a five by five board we substitute n for (m-4)
Where m = board size.
Ym = 2 (m-4)+3
To prove this formula I used the 2nd term.
Ym = 2 (6-4)+3
=7
it can be seen from my results that on a m = 6x6 board there are 7 1x5 rectangles.
When overlapping rectangles of a certain size are counted on different sized boards a quadratic sequence is found. I have used a 1x4 rectangle for this example.
RESULTS
4x4 board
1x4=8
5x5 board
1x4=20
6x6 board
1x4=36
7x7 board
1x4=56
8x8 board
1x4=80
From these results a quadratic sequence can be found.
Using the general term for a quadratic sequence
Yn = An +Bn + C
Where A = ½ 2nd difference = ½ x 4 = 2
A=2
Yn = 2 n + Bn + C
So y1 = 2 x 1 + B x 1 + C = 8
= 2+B+C=8
=B+C=8-2
=B+C=6
so y2 = 2 x 2 + B x 2 + C = 20
8 + 2 B + C = 20
2B+C=20-8
2B+C=12
Subtract the simultaneous equations
2B+C = 12
B+C = 6
B = 6
C=0
If B = 6 and C = 0 the formula is
Yn = 2n +6n
Because we have started form 4x4 board this is the first term of the sequence and so n is substituted for m-3
Ym = 2 (m-3) +6 (m-3)
Ym = 2 (m-3) +6m-18
So for a 5x5 board
Y2 = 2 (5-3) +6 x 5 – 18
= 8 +12
=20
there are 20 1x4 rectangles on a 5x5 board so the formula is proven.
By counting the total number of squares from all the boards, ranging from a 1x1 board to a 8x8 board, and tabulating them, a cubic sequence is found, i.e. the third difference is constant.
Using the general term for a cubic sequence, where A = 1/6 of the 3rd difference
A = 1/6 x 2 = 1/3
Yn = An + Bn + Cn + D
Y1 = 1/3 1 + B x 1 + C x 1 +D
= 1/3 + B + C + D = 1
= B + C + D = 1 – 1/3
= B + C + D = 2/3
y2 = 1/3 x 2 + B x 2 + C x 2 + D
= 1/3 x 8 + B x 4 + 2C + D
= 2 2/3 + 4B + 2C + D = 5
= 4B + 2C + D = 5 – 2 2/3
= 4B + 2C + D = 2 1/3
y3 = 1/3 x 3 + B x 3 + C x 3 + D
= 1/3 x 27 + 9B + 3C + D = 14
= 9B + 3C + D = 14 – 9
= 9B + 3C + D = 5
Solve the simultaneous equation
- B + C + D = 2/3
- 4B + 2C + D = 2 1/3
- 9B + 3C + D = 5
Subtract 2 and 1 = 3 B+C = 1 2/3 4
Subtract 3 and 2 = 5 B+C = 2 2/3 5
Subtract 4 and 5 = 2B = 1
Therefore B = ½
Insert B = ½ into equation 3B+C = 1 2/3 to find C
3x ½ + C = 1 2/3
= 1 ½ + C = 1 2/3
C = 1 2/3 – 1 ½ = 1/6
C = 1/6
To find D insert B = ½ and C = 1/6 into
B+C+D = 2/3
= ½+1/6+D = 2/3
½+1/6 = 2/3
so D = 0
The formula for this cubic sequence is
Yn = 1/3 n + 1/2n + 1/6n
Solve for the first term
Y1 = 1/3 x 1 + ½ x 1 + 1/6 x 1
= 1/3 + ½ + 1/6
= 1
Solve for the fifth term
Y5 = 1/3 x 5 + ½ x 5 + 1/6 x 5
= 41 2/3 + 12 ½ + 5/6
= 55
Predict the ninth term
Y9 = 1/3 x 9 + ½ x 9 + 1/6 x 9
= 243 + 40 ½ + 1 ½
= 285
Predict the tenth term
Y10 = 1/3 x 10 + ½ x 10 + 1/6 x 10
= 333 1/3 + 50 + 1 2/5
= 385
Y9 and y10 can now be put into the table of results and shown to be correct, i.e. the third difference is still constant
