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GCSE: Number Stairs, Grids and Sequences

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  1. Number Grid.

    3 by 3 boxes 4 by 4 boxes. 5 by 5 boxes Results: Term (n) 1 2 3 4 5 Differences in opposite corners 0 10 40 90 160 Term (n)...... 1 2 3 4 5 Differences in Opposite corners...... 0 10 40 90 160 10 30 50 70 20 20 20 We know that the nth term will have to be squared.

    • Word count: 471
  2. Diagonal Difference

    I will then try to find a formula which relates to the diagonal difference of each square, I will then further this investigation by trying to find the diagonal difference of a nine by nine grid and a ten by ten grid and find the formula and see if it is the same. I will also do an extension by doing a rectangle instead of a square and then find the diagonal difference and the formula for this. I am going to find the formula by finding the diagonal difference of all the sizes within the eight by eight grid, and then try to find any patterns, which would help me in finding the formula by drawing a grid.

    • Word count: 1160
  3. Number grid

    The 10 that is underlined in the above equation, proves that the latter part of the equation will always be 10 more than the first half of the equation. So, by using a 2x2 grid and multiplying the diagonals, the difference will always be 10. I am now going to investigate whether there is any relationship between the products of the diagonals of a 3x3 grid. 42 43 44 52 53 54 62 63 64 78 79 80 88 89 90 98 99 100 6 7 8 16 17 18 26 27 28 18 19 20 28 29 30 38

    • Word count: 6538
  4. The Open Box Problem

    This shows that the maximum volume is obtained when x is between 1 and 1.5, I will now make another three spreadsheets, to find the maximum volume correct to three decimal places, as I believe that this first spreadsheet does not give an accurate result. These show, that for a square of width 6cm, that the maximum volume is obtained when x=1.000. I will now repeat the spreadsheets above for a square of width 12cm, to find the maximum volume for this size of card, and to attempt to find any patterns in maximum volumes.

    • Word count: 2831
  5. First Problem,The Open Box Problem

    1cm by 1cm, piece of square card. Length of the section (cm) Height of the section (cm) Depth of the section (cm) Width of the section (cm) Volume of the cube (cm3) 0.1 0.1 0.8 0.8 0.064 0.2 0.2 0.6 0.6 0.072 0.3 0.3 0.4 0.4 0.048 0.4 0.4 0.2 0.2 0.016 0.5 2cm by 2cm, piece of square card Length of the section (cm) Height of the section (cm) Depth of the section (cm) Width of the section (cm)

    • Word count: 1826
  6. To investigate the patterns generated from using rules in a square grid

    To prove this I will let "n" be........................... Therefore using the diagram below the corners are.... This gives......................which proves............. Algebraic Proof 2 If I were to consider rectangles then................. I could prove this by................................. Extension I decided to use cuboids in three dimensions by....... I therefore found........................................... Final Conclusion In this project I found that............................ I have successfully...................................... If I were to extend this project further I would........ Appendices Bibliography Internet print outs Rough Work Maths Coursework Opposite Corners Aim: My task is to investigate the differences of the products of the diagonally opposite corners of a rectangle, drawn on a 10x10 grid, with the squares numbered off 1 to 100.

    • Word count: 1861
  7. For this coursework - stair shape - I am going to investigate the relationship between the stair total and the position of the stair shape on the grid. To do this I am going to create tables, charts, graphs, algebra equations and try to find the n'th ter

    I also plan to show my working outs, and my method of my data capture. Once I have done this I will predicate the n'th term then test to see if my formula works. Data This data that I have collected was from a 10, by 10 grids. The data is presented in a table below. Number inside stair Shape Stair Total 1+2+3+11+12+21 21 11 12 1 2 3 50 6 2+3+4+12+13+22 56 22 12 13 2 3 4 6 3+4+5+13+14+23 62 23 13 14 3 4 5 6 4+5+6+14+15+24 68 24 14 15 4 5 6 *Number in Red is the different that the Stair total is going up in.

    • Word count: 672
  8. I am going to investigate taking a square of numbers from a grid, multiplying the opposite corners and finding the difference of these two results

    So I can see like in the 5x5 grid there is a pattern. If I am right every 2x2 square in a 6x6 grid should have a difference of 6. To check if I am right I will take one more square out of the grid. 16 17 22 23 16 x 23 = 368 17 x 22 = 374 374 - 368 = 6 This shows that I am right and every 2x2 square in a 6x6 grid will have a difference of 6.

    • Word count: 5357
  9. I will attempt to discover a general formula that will find the difference between the product of the top left number and the bottom right and the product of top right and bottom left (diagonals) of any size square and any size grid.

    x 55 = 2420 45 x 54 = 2430 Difference = 10 77 78 87 88 77 x 88 = 6776 78 x 87 = 6786 Difference = 10 18 19 28 29 18 x 29 = 522 19 x 28 = 532 Difference = 10 Pattern: As you can see, the difference between the product top left number with the bottom right and top right and bottom left is 10 every time, to prove this we need to find an algebraic formula.

    • Word count: 1864
  10. How many squares in a chessboard n x n

    =1 square. GENERATING DATA AND DESCRIPTION OF STRUCTURE Applying the above information, we decided to generate the data starting from the smallest square, which is a 1 x 1 square to the largest square, which is n x n. This is a one by one square and can be represented mathematically as 12 (1 squared) or (n - 7)(n - 7) =1 x 1 = 1, where n = 8 This is a 2 by 2 square plus the whole square which consists of 4 squares.

    • Word count: 1890
  11. Investigation - stair shape on a 10x10 numbergrid

    I added up all the numbers in the stair shape, using a calculator, to ensure the results are accurate... The answer = 50 Therefore; Stair no. = 1 Stair total = 50 I proceeded to apply this method to the next few stair no.s To be able to compare the relationship between stair no. and total horizontally across the grid, I am going to chose the next four stair no.s to the right of 1.i.e stair no.s 1-5.This will make the formulae easier to work out.I found the following results; Stair no. 1 2 3 4 5 Stair total 50 56 62 68 72 74 +6 +6 +6 +4?

    • Word count: 1356
  12. Number Grids.

    After doing this I wondered if this theory would work if I used a 2*3 rectangle. 27*48=1296 28*47=1316 Diagonal difference=1316 - 1296=20 27 28 37 38 47 48 34*55=1870 35*54=1890 Diagonal difference=1890 - 1870=20 34 35 44 45 54 55 So I then from this I wondered if larger rectangles had the same diagonal difference from this I found: - Rectangle Rows * columns Diagonal difference 2*3 20 3*4 60 4*5 120 5*6 200 6*7 300 7*8 420 From this I will try to find a formula: - Rectangle Rows * columns Diagonal difference This goes up in tens (multiples of ten).

    • Word count: 1987
  13. This is a 3-step stair. The total of the numbers inside the step square shape is 25+26+27+35+36+45=194. The total for this step square is 194.

    This formula can be used to find out the total of that the Step Square using the bottom left hand corner. We can prove this with an example. If the bottom left hand corner number were 4 the formula would be 6 x 4+44= 68 or without the formula 4+5+6+14+15+24=68. This proves that the method works for a three-step formula. Part 2 Aim : The aim of the next part of the project is to find out what a 3x3 4x4 5x5 or any other step square would be in a 10x10 grid.

    • Word count: 1886
  14. Number Grids Investigation

    After doing this I wondered if this theory would work if I used a 2*3 rectangle. 27 28 37 38 47 48 34 35 44 45 54 55 So I then from this I wondered if larger rectangles had the same diagonal difference from this I found: - Rectangle Rows * columns Diagonal difference 2*3 20 3*4 60 4*5 120 5*6 200 6*7 300 7*8 420 From this I will try to find a formula: - Rectangle Rows * columns Diagonal difference 2*3 20 2*4 30 2*5 40 2*6 50 2*7 60 2*8 70 .�. As the number of columns goes up a further 10 is added so I predict a 2*9 rectangle will be 80.

    • Word count: 1721
  15. Number Grids

    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 From the squares I can find the difference for a 2x2 grid.

    • Word count: 1973
  16. Number Grid Investigation

    the top right and bottom left numbers: 24*33=792 Next I found the difference: 792-782=10 I then did this again for another square: 36 37 46 47 36*47=1692 37*46=1702 1702-1692=10 I did this a few times and kept on getting ten as the difference. I then worked out if the difference was always ten by working out what the value of each corner in the square is relative to the top left square. X X+1 X+10 X+11 ((X+1)(X+10))-(X(X+11))=Difference (X�+10X+X+10)-(X�+11X)=Difference (X�+11X+10)-(X�+11X)=Difference Difference = 10 This shows that the difference with a 2 by 2 square is always ten when it is in a grid with ten squares in each row.

    • Word count: 1015
  17. Borders Maths Coursework

    With this shape there are now 12 borders. This shape has 16 borders I can now see a pattern in the number of square, which is that there are 4 more each time.

    • Word count: 279
  18. Corners - Maths Investigation

    RULE d = c I shall extend this by varying the size of the square extracted from the grid. The first stage of this will be to give an algebraic value to each . n n+1 n+2 n+c n+c+1 n+c+2 n+2c n+2c+1 n+2c+2 I shall begin my working by extrapolating the difference between products, as before. n(n+2c+2) = n�+2cn+2n (n+2)(n+2c) = n�+2cn+2n+4c n�+2cn+2n+4c - (n�+2cn+2n) n�+2cn+2n +4c -(n�-2cn-2n) 4c I currently have insufficient data to begin searching for a rule, so I shall repeat the above process with a 4x4 square. n n+1 n+2 n+3 n+c n+c+1 n+c+2 n+c+3 n+2c n+2c+1 n+2c+2 n+2c+3 n+3c n+3c+1 n+3c+2 n+3c+3 n(n+3c+3)

    • Word count: 1550
  19. Maths Grid Investigation

    What I think will happen I think that the diagonal difference will always be 32 with a 3x3 grid. I think that the diagonal difference for a 2x2 grid will be 8 each time. I am also going to investigate the diagonal difference of a grid sized 4x4; I think the difference will be 72 each time. I came to these conclusions by doing a series of preliminary investigations. Preliminary Investigations For a 3x3 grid: 1 2 3 9 10 11 17 18 19 1 x 19 = 19 3 x 17 = 51 51 - 19 = 32 46

    • Word count: 915
  20. China in the 20th Century Sources Questions

    In the past they have just stood back and watched as dissidents' wives stimulated publicity." I think that this is an advantage to the spouses, as the women would just make a fool of themselves by stimulating publicity. However, they would have had an effective response by stimulating publicity because then other citizens would know what is going on. "In a similar case, the dissident w**g Dan, a leader of the 1989 Tiananmen Square movement, has gone into hiding after receiving death threats police who stalked him for more than a year and a half.

    • Word count: 2643
  21. Number Grid Product Differences Investigation

    - (73x82) = 10. This shows that the product difference is the same anywhere in the grid. Then we were told that we had to find a general formula to work out the product difference in 10x10 grid but with a 3x3 or 5x5 matrices inside. I firstly found out the product difference in a 3x3, 4x4, 5x5 and a 6x6 grid. Below are the differences of the matrices. Here is another 2 more examples to prove that the formula above works 74 75 84 85 I did (74x85) - (75x84) = 10. This shows that the product difference is the same anywhere in the grid. 59 60 69 70 I did (59x70) - (60x69) = 10.

    • Word count: 3124
  22. Maths Grid Investigation

    grids of all sizes not yet done Page 26: Justifying My Results, formula tested on grids of all sizes not yet done Page 27: Further Investigation, formula found for any rectangle grid inside any square grid. Page 28: Justifying My Results, formula tested on grids of all sizes not yet done Page 29: Justifying My Results, formula tested on grids of all sizes not yet done Page 30: Conclusion Statement I have been told to consider the following table of numbers: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

    • Word count: 5120
  23. A garden centre sells square slabs which can be used to surround ponds

    To work out the perimeter of a square the formula is 4n. Using this 4x1=4. I need 8 slabs so that it goes all the way around the pond. I need four more. That makes 8. I will try using the formula 4n+4 to see how many slabs are needed. It works for a square when the sides are 2x2 and 3x3. 4x2 = 8. 8+4 = 12 4x3 = 12. 12+4 = 16 The amount of slabs are correct.

    • Word count: 528
  24. Number Grid Investigation

    I will then work out formulas for both of these grids, to find the difference. I will then use algebra to prove that my formulas are able to work out the correct difference. To investigate further I will do the whole investigation again but with rectangles instead of squares. This is a table to show the differences in squares in a 10 by 10 grid. No. of Rows (r) No. of Columns (c) Difference Formula 2 2 10 10(r-1)� 3 3 40 10(r-1)� 4 4 90 10(r-1)� 5 5 160 10(r-1)� Firstly, I noticed that if you took the zeros off the difference, then you would get square numbers for 1, 2, 3 and 4.

    • Word count: 1696
  25. Routes On Polyhedra

    at G. My method for the cuboid in the diagram above is to start at A (*) and then find all the routes that began A-B, then all the routes that began A-D and finally A-E.

    • Word count: 283

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