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  • Level: GCSE
  • Subject: Maths
  • Word count: 1873

Block sequences

Extracts from this document...

Introduction

Block sequences

The aim of this project is to investigate a sequence of blocks. I am going to try to find a pattern (Nth term). I will place the results of my findings alongside results from other shapes in an attempt to find a general pattern for the areas that I have looked at

        The areas that I will investigate are: The perimeter, The area, The number of unshaded squares and The number of squares.

        I will begin my investigation with the sequence that I was given.

Through my investigation I will

Replace

With

Perimeter

P

Area

A

Shaded

S

Unshaded

Uns

Squares

P

4

A

1

S

0

Uns

1

image00.png

P

12

A

5

S

4

Uns

1

image01.png

P

20

A

13

S

8

Uns

5

image03.png

P

28

A

25

S

12

Uns

13

image04.png

P

36

A

41

S

16

Uns

25

image05.png

I will now put all the figures I have recorded into tables and find the Nthterm

Perimeter

Diagram

No of sides

1

4

2

12

3

20

4

28

5

36

4

12

20

28

36

    8

    8

    8

    8

Now that I have found the common difference in the sequence I can work out the Nthterm.

Diagram

8N

-

=

Perimeter

1

8

4

=

4

2

16

4

=

12

3

24

4

=

20

4

36

4

=

28

5

40

4

=

36

The Nth term for the perimeter is 8N- 4

I will now follow the same steps to work out the Nth term for the Area, number of shaded and the number of unshaded.

Area

Diagram

Number of squares

1

1

2

5

3

13

4

25

5

41

1

5

13

25

41

    4

    8

    12

    16

4

4

2A = 4     A = 2                         Nth

...read more.

Middle

1

5

13

25

    0

    4

    8

    12

4

4                        4

2A = 4     A = 2                         Nth term = 2N2 - 6N + 5

3A + B = 0

6 + (-6) = 0

A + B + c = 1

2 + (-6) + 5 = 1

To make sure that I have worked out all my equations properly I am now going to draw a test diagram to make sure. I predict that the 6th diagram in the sequence will have.

Perimeter

8N - 4

(8x6) - 4

= 44

Area

2N2 - 2N + 1

2(6x6) - (6x2) + 1

= 61

Shaded

4N - 4

(6x4) - 4

= 20

Unshaded

2N2 - 6N + 5

2(6x6) - (6x6) + 5

= 41

N = 6 because it is the 6th diagram, if it was the 7th diagram then N = 7 and so on.

P

44

A

61

S

20

Uns

41

image06.png

I am now going to duplicate my results for another shape (triangles) to see If the sequences are the same, if not I will look for a pattern between them, a general term

Triangles

P

3

A

1

S

0

Uns

1

image07.png

P

6

A

4

S

3

Uns

1

image08.png

P

12

A

10

S

6

Uns

4

image09.png

P

15

A

19

S

9

Uns

10

image10.png

P

21

A

31

S

12

Uns

19

image02.png

I will now put all the figures I have recorded into tables and find the Nthterm

Perimeter

Diagram

No of sides

1

3

2

6

3

12

4

15

5

21

3

6

12

15

21

    3

    6

    3

    6

3

-3

3

I have noticed that on the 2nd difference the numbers are repeating (3, -3, 3, -3….) because of this there is no way to be sure that further on down in the differences that it will ever be the same. I have found that by using 2 Nthterms 9(N/2 + 0.5) - 6 and 9(N/2)  - 3

Diagram

Odd

Even

=

Perimeter

1

9(1/2 + 0.5)  - 6

=

3

2

9(2/2) - 3

=

6

3

9(3/2 + 0.5) - 6

=

12

4

9(4/2) - 3

=

15

5

9(5/2 + 0.5) - 6

=

21

Area

Diagram

No of triangles

1

1

2

4

3

10

4

19

5

31

1

4

10

19

31

    3

    6

    9

    12

3

3

3

For the sequences in triangles I will use simultaneous equation to work out the Nth term.

  A + B + c = 1    ---------(1)

4A + 2B + C = 4 ---------(2)

9A + 3B + C = 10 --------(3)

To work this equation out I will have to eliminate one of the letters. I will remove the +c because it is in all the equations

(2) - (1)

(2) - (1)

4A + 2B + C = 4

9A + 3B + C = 10

A + b + C = 1

A + B + C = 1

3A + B = 3-------- (4)

8A + 2B = 9------------(5)

((4) X 2) - (5)

6A + 2B = 6

8A + 2B = 9

-2A = -3                  A = 1.5

...read more.

Conclusion

5th Diagram

Area - 3N2 - 3N + 1

    (3 X 52) - (3 X 5) + 1

   (3 X 25) - (3 X 5) + 1

       75 - 15 + 1 = 61

Shaded - 6N - 6

           (5 X 6) - 6

              30 - 6 = 24

Unshaded - 3N2 - 9N + 7

            (3 X 52) - (9 X 5) + 7  

            (3 X 25) - (9 X 5) + 7

                    75 - 45 + 7 = 37

Perimeter - I was unable to find a general term for the perimeter so I will work it out

6

18

30

42

    12

    12

    12

General term for the perimeter is 12N - 6

Conclusion

During this investigation I have looked at different ways of solving numeric sequences derived from patterns found in shapes. From the sequence I was originally given I have worked out Nth terms that help describe the patterns in the shape (Perimeter, Area, Number of shaded and the Number of unshaded). I moved on to another shape (triangles) to see if the pattern continued similarly in other tessellating shapes. I found that the patterns might not have been the same, but by dividing the sequences by the number of sides in each category, I could get a general Nth term. This general Nth term can be used to find the pattern in any shape simply by multiplying the general term by the number of sides in the shape you are investigating. I looked to see if there was a single term that could be used to describe the pattern in all shapes but I was unable to find it.

...read more.

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