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  • Level: GCSE
  • Subject: Maths
  • Word count: 1416

Investigation to find out the number of matchsticks on the perimeter in a matchstick staircase using the GENERAL RULE.

Extracts from this document...

Introduction

Mary Anitha Edward Antony

Matchsticks Coursework

Introduction

This investigation is based on the ‘number sequence’ and I am going to make further more matchstick staircases for this investigation.

Investigation to find out the number of matchsticks on the perimeter in a matchstick staircase using the GENERAL RULE.

I have drawn 6 matchstick staircases on the graph paper and I am going to put the number of matchsticks on the base, number of matchsticks on the perimeter, total number of matchsticks in a table based on the 6 matchstick staircases.

Table to show the number of matchsticks on the base, on the perimeter and the total number of matchsticks.

Number of matchsticks on the base

Number of matchsticks on the perimeter

Total number of matchsticks

1

4

4

2

8

10

3

12

18

4

16

28

5

20

40

6

24

54

And I’m going to make another table to find out the differenceonperimeter from the number of matchsticks on the perimeter.

Number of matchsticks on base

1

2

3

4

5

6

Number of matchsticks on perimeter

    4            8           12          16          20          24

           4            4            4            4            4

Perimeter difference

From this table I’m going to make a general rule, in terms of letters.

Number of matchsticks on perimeter = P

Number of base = b

Perimeter difference = 4 (always)

...read more.

Middle

1st difference

2nd difference

When the 1st difference is not the same but the 2nd difference is, the formula will follow the quadratic pattern:

t = an2 + bn + c

But I am going to use different letters.

t = total; b = base

t = xb2 + yb + c

x is the coefficient of the 1st term and is always ½ the 2nd difference.

So in my example x is ½ of 2 = 1.

∴My formula will begin with 1b2 = b2.

To find ‘c’ I have to find the value of t, when b = 0. So I am going draw another table.

B

0

1

2

3

4

5

6

t

    0            4          10          18          28         40         54

4            6           8           10          12         14

                  2            2            2            2           2

1st difference

2nd difference

In my table when b = 0, t = 0 so c = 0.

Now I have to find the value of ‘y’.

x + y + c = the value of t when b = 1 (from the table)

x + y + 0 = 4

1 + y + 0 = 4

y = 3

So my formula should be:

b2 + 3b = t

Now I am going to test my formula:

Example 1: when b = 6; t = 54

b2 + 3b = t

62 + 3(6) = 54

36 + 18 = 54

I am going to try another example to make sure whether my formula is right or not.

Example 2: when b = 2; t = 10

b2 + 3b = t

22 + 3(2) = 10

4 + 6 = 10

Therefore I would say my formula is right and using this formula I will predict that the total number of matchsticks in the diagram when the base is 9 is:

b2 + 3b = t

92 + 3(9) = t

81 + 27 = 108.

Introduction

I am going to do another investigation to find out the matchsticks on the perimeter and the total number of matchsticks in a double matchstick staircase. I have drawn four diagrams (see the graph attached).

Investigation to find out the number of matchsticks on the perimeter in a double matchstick staircase using the GENERAL RULE.

Table to show the number of rows, number of matchsticks on perimeter and the total number of matchsticks.

r = number of rows; P = number of matchsticks on perimeter; t = total number of matchsticks

r

P

t

1

4

4

2

10

13

3

16

26

4

22

43

...read more.

Conclusion

t = ar2 + br + c

But I am going to use different letters.

t = total; r = rows

t = xr2 + yr + c

x is the coefficient of the 1st term and is always ½ the 2nd difference.

So in my example x is ½ of 4 = 2.

∴My formula will begin with 2r2.

To find ‘c’ I have to find the value of t, when r = 0. So I am going draw another table.

R

0

1

2

3

4

t

   -1           4          13          26          43        

5            9          13         17          

                 4            4           4      

1st difference

2nd difference

In my table when r = 0, t = -1 so c = -1.

Now I have to find the value of ‘y’.

x + y + c = the value of t when r = 1 (from the table)

x + y + -1 = 4

2 + y + -1 = 4

y = 3

So my formula should be:

2r2 + 3r -1 = t

Now I am going to test my formula:

Example 1: when r = 3; t = 26

2r2 + 3r -1 = t

2(3)2 + 3(3) -1 = 26

2(9) + 9 -1 = 26

18 + 9 -1 = 26

I am going to try another example to make sure whether my formula is right or not.

Example 2: when r = 4; t = 43

2r2 + 3r -1 = t

2(4)2 + 3(4) -1 = 43

2(16) + 12 -1 = 43

Therefore I would say my formula is right and using this formula I will predict that the total number of matchsticks in the diagram when the rows are 9 is:

2r2 + 3r -1 = t

2(9)2 + 3(9) -1 = t

2(81) + 27 -1 = 188.

...read more.

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