• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

rectangles. I will be trying to develop a formula that will enable me to calculate the sum of all the numbers in a rectangle given

Extracts from this document...

Introduction

image00.png

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

46

47

48

49

50

51

52

53

54

55

56

57

58

59

60

61

62

63

64

65

66

67

68

69

70

71

72

73

74

75

76

77

78

79

80

81

82

83

84

85

86

87

88

89

90

91

92

93

94

95

96

97

98

99

100

image01.png

image02.png

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

46

47

48

49

50

51

52

53

54

55

56

57

58

59

60

61

62

63

64

65

66

67

68

69

70

71

72

73

74

75

76

77

78

79

80

81

82

83

84

85

86

87

88

89

90

91

92

93

94

95

96

97

98

99

100

The value of the rectangle is 132.

We would get the value by calculating the sum of all the numbers.

A way we could find this value out could be by using a formula. But what formula could we use?

n

n+1

n+2

n+10

n+11

n+12

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

46

47

48

49

50

51

52

53

54

55

56

57

58

59

60

61

62

63

64

65

66

67

68

69

70

71

72

73

74

75

76

77

78

79

80

81

82

83

84

85

86

87

88

89

90

91

92

93

94

95

96

97

98

99

100

The sum of the values in the turquoise box adds up to 132.

To find this, we could try n and n+12.

16+16+12=44

132÷44=3

3 are equal to the width.

So far we have w (2n + 12).

...read more.

Middle

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

46

47

48

49

50

51

52

53

54

55

56

57

58

59

60

61

62

63

64

65

66

67

68

69

70

71

72

73

74

75

76

77

78

79

80

81

82

83

84

85

86

87

88

89

90

91

92

93

94

95

96

97

98

99

100

This time, the width of each shaded square is 4.

For the first shaded square, the values come up to 180.

4(2n+12): 4(44): 4x44= 176.

As you can see, we have changed the width, but have not changed the n+12 to n+13. If we tried this, we would get 180.

To do this so our formula is correct we would have to alter it.

W (2n+r). ‘R’ stands for the range. To make sure this formula is correct, I will try it on the remaining two squares and then on one with a completely different width.

The sum of the rose coloured box comes up to 380.

...read more.

Conclusion

R(LxW)+(LxW)

      2

14(2x5)+(2x5)

2

14(10): 140. 140÷2=70. 70+(10)=80. So far this formula has been correct, but would happen if initial term changes from 1 to a different number.

R(LxW)+(LxW)

      2        

12(2x3)+(2x3)

2

12(6)=72. 72÷2=36. 36+6=42.

This answer is incorrect.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

46

47

48

49

50

51

52

53

54

55

56

57

58

59

60

61

62

63

64

65

66

67

68

69

70

71

72

73

74

75

76

77

78

79

80

81

82

83

84

85

86

87

88

89

90

91

92

93

94

95

96

97

98

99

100

I could now try to split the height from 2 to 1 and adding both rows to get the final product.

w(in+1). I can change in which is the initial term to n, which will stand for the lowest number.

Width(n+1). 3(4+1)=3(5)=15. 4+5+6=15.

For the second row we will have 3(14+1)=3(15)= 45. 14+15+16=45.

45+15+60.

4+5+6+14+15+16=60.

54+55+56+57+64+65+66+67=484.

4(54+1) 4(55)= 220.

4(64+1) 4(65) 195. 220+260=480. This answer is incorrect. So instead of it being w(n+1), I am going to change it to w(n+(r-1)) for each row.

4 (54+(2))= 4(56)= 224

4(64+2)=4(66)=264. 224+264= 488. this answer is also incorrect.

...read more.

This student written piece of work is one of many that can be found in our GCSE Number Stairs, Grids and Sequences section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Number Stairs, Grids and Sequences essays

  1. GCSE Maths coursework - Cross Numbers

    X (X+1) X+g c) [(X+g) - (X-g)] - [(X+1) - (X-1)] = [X+g - X + g] - [X+1 - X +1] = x+g - x + g -x -1+ x - 1 = 2g - 2 Proving the above formulas with a grid size 10x10 X-g (X-1) X (X+1)

  2. Open Box Problem.

    of x, which gives this open box its maximum value, is 5. Notice that even though the ratio has gone form 1:3 to 1:4, the length (which is 20cm for both ratios) divided by 4 still gives the cut of x that gives this open box its maximum volume.

  1. Investigate the difference between the products of the numbers in the opposite corners of ...

    1 2 3 11 12 13 21 22 23 1 x 23 = 23 3 x 21 = 63 63 - 23 = 40 8 9 10 18 19 20 28 29 30 8 x 30 = 240 10 x 28 = 280 280 - 240 = 40 I can

  2. I will try to find a formula linking P (perimeter), D (dots enclosed) and ...

    dots with many other triangles, therefore there are much more dots enclosed than if the triangles were laid in a line 10 Triangles (T=10): P= D= T= 8cm 2 10 10cm 1 10 12cm 0 10 15 Triangles (T=15): P= D= T= 11 3 15 13 2 15 15 1

  1. Investigate the relationships between the numbers in the crosses.

    - (d+11)(d-9) = (d +2d) - (d -9d+11d-99) = (d +2d) - (d +2d-99) = -99 * If x is known: = (x-1)(x+2) - (x+10)(x-10) = (x +1x-1x-1) - (x -10x+10x-100) = (x +x-1) - (x +x-100) = 99 * This also implies that the outcome of the sum (d x b)

  2. My investigation will be on 3 - step stairs where I will be: ...

    26 27 28 29 30 11 12 13 14 15 16 17 18 19 20 1 2 3 4 5 6 7 8 9 10 21 11 12 1 2 3 22 12 13 2 3 4 23 13 14 3 4 5 24 14 15 4 5 6 25

  1. For my investigation I will be finding out patterns and differences in a number ...

    x 6 = 306 306 - 56 = 250 45 46 47 48 49 50 55 56 57 58 59 60 65 66 67 68 69 70 75 76 77 78 79 80 85 86 87 88 89 90 95 96 97 98 99 100 45 x 100 = 4500

  2. The Open Box Problem

    These show that the maximum volume for a square of width 12cm is obtained when x=2. Width of square Value of x to give maximum volume 6 1 12 2 I have noticed that so far, my results would suggest that the maximum volume if given when x is 1/6th of the width of the square sheet of card.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work