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The figure below

Paper type: Essay
Pages: 4 (878 words)
Categories:
Downloads: 36
Views: 1

The figure below shows a dark cross-shape that has been surrounded by white squares to create a bigger cross- shape:

The bigger cross- shape consists of 25 small squares in total. The next cross- shape is always made by surrounding the previous cross-shape with small squares.

This coursework will be to investigate to see how many squares would be needed to make any cross-shape build up in this way.

Method

First of all I’m going to work out a sequence which consists of a formula.

Then I will draw a maximum of six cross-shapes adding on to each sequence, at the end I will produce one extra to testify the formula.

After creating two tables to work out the first and second differences between the number of squares, I’ll use those to find the values of a , b and c.

Table 1

Shape Number

1

2

3

4

5

6

Number of Squares

1

5

13

25

41

61

1st Difference

4

8

12

16

20

24

2nd Difference

4

4

4

4

4

4

No of squares:

1 5 13 25 41 61

1st Diff:

4 8 12 16 20

2nd Diff:

4 4 4 4

To find the 1st and 2nd difference, I used a table and then the diagram above to check if I was right about my suggested answers.

First of all, for the 1st difference, I took the number of squares for shape number 2 and subtracted that with the number of squares I got for the first shape, which then gave me the 1st difference. I continued like this until I had all my 1st differences.

Afterwards, I took the result from the 1st difference of shape number 2 and subtracted that with the 1st difference of shape number 1, I repeated this until the end, and the 2nd difference turned out to be 4. This is called a quadratic function, this means when it takes two sets of terms to actually find out what the constant difference is.

In the following table I will be intending to find out what the answers are to 2n2.

Table 2

Shape Number

1

2

3

4

5

6

Number of Squares

1

5

13

25

41

61

2n2

2

8

18

32

50

72

Remaining sequence

-1

-3

-5

-7

-9

-11

1st difference

-2

-2

-2

-2

-2

-2

No of squares:

1 5 13 25 41 61

Remaining Sequence:

-1 -3 -5 -7 -9

1st Diff:

-2 -2 -2 -2

To find the remaining sequence, I used the formula 2n2 with the number of squares . So basically when I worked out the answer to 2n2, I subtracted the number of squares by the results of my formula, which gave me the answers to the remaining sequence.

2 12 = 2

2 52 = 8

2 132 = 18

2 252= 32

2 412= 50

2 612= 72

From there, to get the 1st differences, I took the remaining number of the second shape and subtracted that to the remaining number of the first shape. I continued until I had all the same results which was -2.

The formula I was to investigate was:

an + bn + c

This would become:

2n + -2n + 1

To testify this formula I will draw a 7th cross-shape to test that:

n = 7

This would therefore become:

2 ( 7 ) – 2 ( 7 + 1 ) = 85

At the end I’ll make sure that I drew the right amount of squares for my shape by counting appropriately.

Table 3

Shape Number

1

2

3

4

5

6

No of Squares

7

25

63

129

231

377

1st Diff

18

38

66

102

146

2nd Diff

20

28

36

44

3rd Diff

8

8

8

As I have had 3 sets of differences( this is called a Cubic Function which means anything to the power of 3 ) before the differences remained constant, I now know that my formula will n3 in it.

The 3rd difference is therefore 8 which means:

8 a= 8

6

The 1st part of the formula becomes 8 nb = 4 n3

6 3

I will now use 4 n2 to make a sequence.

3

n

1

2

3

4

5

6

4

3 n3

4

3

32

3

158

3

256

3

500

3

864

3

Remaining Sequence

=

1 – 4

3

-1/3

7 – 32

3

-11/3

25- 158

3

-33/3

63 – 256

3

-67/3

1st Diff

-10

3

-22

3

-34

3

2nd Diff

-12

3

-12

3

-12

3

The second part of the equation will now be:

-6 n2 = -2n2

3

I will use this to make another sequence.

n

1

2

3

4

5

6

-2n2

-2

-8

-18

-32

Remaining Sequence

-1/3-(-2)

5/3

-11/3-(-8)

13/3

-33/3-(-18)

21/3

-67/3-(-32)

29/3

37/3

1st Diff

8/3

8/3

8/3

8/3

I have now found out the last part of the formula which is :

8 n – 3

3 3

The formula is : 4/3n3 – 2n2 + 8/3n – 3/3

= 4n3 – 6n + 8n – 3

3

To testify this formula I’m going to draw squares so that the 3D look can be easily understood.

Cite this essay

The figure below. (2020, Jun 02). Retrieved from https://studymoose.com/the-figure-below-new-essay

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