Borders and squares

From the patterns I have carried out I have noticed:

> That this pattern is a changing difference. Because the first difference changed down the column, but the second difference is the same all the way along.

> I can see that the second difference is a square number. Because 2x2=4

Why is the second difference square numbers?

Because the numbers in the first difference in my pattern are even numbers.

Now I am going to use an easy method to try and find my formula:

Because the second difference is +4 we can write it in a form of (22) but since here we want to find a

formula with the nth term we will write it in a form of (2n2)

Then times each of our sequence numbers by it, and then add it to No. Of cubes.

'n' indicates the position in the sequence.

Seq. no

2

3

4

5

6

No. Of cubes

5

3

25

41

61

x(2n2)

2

8

8

32

50

72

Total

-1

-3

-5

-7

-9

-11

In our total answer we can see that the difference between each number is -2:

-1 - 2= -3

-3 -2 = -5

-5 -2 = -7

-7 -2 = -9

-9 -2 = -11

Consequently we will times each of our seq. no by -2n and then add it by our total number in the previous table.

Seq. no

2

3

4

5

6

Total in last table

-1

-3

-5

-7

-9

-11

x-2n

-2

-4

-6

-8

-10

-12

Total

Therefore from these 2 tables we can see that the formula will be (2n2)-2n+1

Now I will draw a table and put both of the tables above in 1 table. That way I will be able to show my work better.

Seq. no

2

3

4

5

6

No. Of cubes

1

5

13

25

41

61

x(2n2)

2

8

8

32

50

72

Total

-1

-3

-5

-7

-9

-11

x-2n

-2

-4

-6

-8

-10

-12

Total

TESTING THE RULE

So now that I have the equation (2n2)-2n+1. I will try to prove my equation by applying it to a number of sequences and higher sequences, which I have not yet explored.

Sequence 3:

. 2(32) - 6 + 1

2. 2(9) - 6 + 1

3. 18 -5

4. = 13

The formula when applied to sequence 3 appears to be

successful.

Sequence 5:

. 2(52) - 10 + 1

2. 2(25) - 10 + 1

3. 50 - 10 + 1

4. 50 - 9

5. = 41

Successful

Sequence 6:

. 2(62) - 12 + 1

2. 2(36) - 12 +1

3. 72 - 12 + 1

4. 72 - 11

5. = 61

Successful

The formula I found seems to be successful, as I have shown above. I will now use the formula to find the number of squares in a higher sequence.

Sequence 12:

. 2 (122) - 24 + 1

2. 2(144) - 24 + 1

3. 288 - 24 + 1

4.800 - 23

5. = 265

To find out if my answer is right or not I am going to see the difference between the numbers from sequence 6 to 20. As we know the difference between each number is +4. I will add the sequence of 6 with its first difference through its previous number, after that I will add it by 4 to get the sequence of 7 and I will do the same with all the other sequences.

Sequence of 7:

61+20+4= 85

Sequence of 8:

85+24+4=113

Sequence of 9:

13+28+4=145

Sequence of 10:

45+32+4=181

Sequence of 11:

81+36+4=221

Sequence of 12:

221+40+4=265

61 85 113 145 181 221 265

24 28 32 36 40 44

+4 +4 +4 +4 +4

From all of these we can see that the formula is right. Because by putting the sequence of 12 in it we obtained the same result as we did above. Therefore I can see that the formula is correct.

CONCLUSION

From all my showings we can see that my prediction was correct. This was a simple way in which I showed how to find the formula.

After investigating the 2D shape I will extend my investigation by making the borders 3 dimensional. This will be a lot harder to investigate because there are 3 dimensions.

I will have to draw the shapes in 2D so I will draw each section from top to bottom.

I am going to draw my first shape and add a new section to the whole.

As you can see above the blue squares are the new section added which is the original shape plus a border. Now there's two of the original shape.

I have draw all the 6 sequences on a paper. So it will be easier for you to see them.

I am first going to find out the number of borders in each sequence. From the borders I have drawn on the paper. Then I will put them in a table.

Seq .no

2

3

4

5

No. of cubs

7

25

63

04

I am going to use the formula below to prove the 3D. by proving this equation, I will hopefully prove myself right.

'n' indicates the position in the sequence.

an4 +bn3 +cn2+dn+e

position of

sequence: an4 +bn3 +cn2+dn+e

n=1 a+b+c+d+e= 1

n=2 16a+ 8b+4c+2d+e=7

n=3 81a+27b+9c+3d+e=25

n=4 256a+64b+16c+4d+e= 63

n=5 625a+125b+25c+5d+e= 129

Now I am going to subtract n=2 from n=1, n=4 from n=3 and n=5 from n=4. Basically I will be subtracting one equation from the equation before it. I will keep on doing this till I get to 2 equations. This way I will be able to find a,b,c,d and e. and then I can go on further to prove the equation.

5a + 7b + 3c+d= 6

65a + 19b+ 5c+d=18

75a +37b + 7c +d=38

369a +61b + 9c +d=66

as you can see above all the Es have been cancelled out because a (-) and a (+) of the same number or letter gets cancelled out.

Again I will do the same as above and subtract each equation from the equation previous it.

50a +12b +2c= 12

10a +18b +2c=20

94a +24b +2c=28

Again in the equations above all the Ds have been cancelled out and that's why I didn't write them down. I'm going to do the same again, this time I will get to 2 equations and so I will be able to find out what number a, b,c,d and e are.

60a +6b= 8

84a +6b= 8

We will subtract these numbers again and the 6bs will be cancelled out and this way we will be able to find a.

24a= 0 a = 0

We will now find b by placing a in the equation:

60 x (0)+ 6b= 8

0 + 6b= 8

6b = 8

b= 4/3

And we will now put a and b in the equation to find c:

50a +12b+ 2c= 12

50 x (0) + 12 (4/3)+ 2c= 12

6+ 2c= 12

2c= 12- 16

2c= -4 c= -2

Now we are going to put a, b and c in an equation to find d:

5a+ 7b+ 3c+ d= 6

5 x (0)+ 7 (4/3) + 3(-2)+d= 6

28/3 - 6+ d= 6

d= 12 - 28/3 d=8/3

I am are going to do the same thing as above to find e. we will place a, b, c and d in the equation to find e:

A+ b+ c+ d+ e= 1

0+ (4/3) - 2+ (8/3)+e= 1

(6/3) + e=1

e= 1 - (6/3) e= -1

Now I am going to draw the diagram for sequence 6 and 10. After counting how many cubes there are I am going to try it on the equation. If I get the same answer then the equation is correct.

The diagrams are on the next page.

Salam