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• Level: GCSE
• Subject: Maths
• Word count: 2734

# Number stairsMy aim is to investigate the relationship between the stair total and the position of the stair shape on the grid for 3 step

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Introduction

Jaie Wilde

Number stairs

My aim is to investigate the relationship between the stair total and the position of the stair shape on the grid for 3 step stairs and to investigate further the relationship between the stair totals and other step stairs on other number grids.

 91 92 93 94 95 96 97 98 99 100 81 82 83 84 85 86 87 88 89 90 71 72 73 74 75 76 77 78 79 80 61 62 63 64 65 66 67 68 69 70 51 52 53 54 55 56 57 58 59 60 41 42 43 44 45 46 47 48 49 50 31 32 33 34 35 36 37 38 39 40 21 22 23 24 25 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

To begin I will try to identify a pattern. Taking my example I will take all the three stair steps on the bottom row and see if there are any similarities between the results.

1. 1+2+3+11+12+21=50
2. 2+3+4+12+13+22=56
3. 3+4+5+13+14+23=62
4. 4+5+6+14+15+24=68
5. 5+6+7+15+16+25=74
6. 6+7+8+16+17+26=80
7. 7+8+9+17+18+27=86
8. 8+9+10+18+19+28=92

Therefore I can now find a formula. The general term for an arithmetic sequence is Un=ab+c. The terms go up in sixes and this tells me that the nth term will include 6 lots of n or 6n.  For the first term n=1, so 6n=6. But the first term is 50 which is 44 more than 6n. This suggests that the formula is 6n+44.

Trying a few values of n will help prove that my formula is correct.

(6 multiplied by 1) +44 =50.

(6 multiplied by 2) +44 =56.

(6 multiplied by 3) +44 =62.

(6 multiplied by 4) +44 =68.

(6 multiplied by 5) +44 =72.

I will now pick a random 3 step stair and test my formula.

 76 66 67 56 57 58

Step 1) Locate n which is in bottom left corner.

N=56

Step 2) Insert n into formula 6n+44.

Step3) (6 multiplied by 56) +44= 380 which equals the sum of the numbers.

The formula has worked.

With this formula I can now identify the sum of any 3 step stairs. I will now go on to try and identify similar formulas for bigger stair steps.

Middle

I will now test my formula by selecting a random 6 step stair.

 72 62 63 52 53 54 42 43 44 45 32 33 34 35 36 22 23 24 25 26 27

I have now successfully found formulas for 3, 4, 5 and 6 step stairs.

 Number of stairs formula 3 6n+44 4 10n+110 5 15n+220 6 21n+385

From this table I have noticed that the numbers before n are triangular. Therefore I assume that the formula for a seven step stair will include 28n.

I will now test this on a seven step stair to see if this is true.

 61 51 52 41 42 43 31 32 33 34 21 22 23 24 25 11 12 13 14 15 16 1 2 3 4 5 6 7

Also I have noticed that the number of stairs corresponds to the first term of the formula.

 3 6n+44 4 10n+110 5 15n+220 6 21n+385

To get from 3 to 6 you multiply by 2.

To get from 4 to 10 you multiply by 2.5.

To get from 5 to 15 you multiply by 3.

To get from 6 to 21 you multiply by 3.5.

If I put this in a table:

 1 2 3 4 5 6 1 1.5 2 2.5 3 3.5

The difference between the terms is ½ so my formula must start with ½ n. If I take term 4 I will half it to get two then I will add half to get to 2.5. This applies to all the terms. Therefore my formula is ½ n+ 0.5. This shows me what to multiply the number of step stairs by to find out the first term of the formula.

I will take an 8 step stair as an example to prove my formula.

If I put 8 in the formula it would be ½ of 8 which is 4. Then add 0.5 which is 4.5. This shows me that I have to multiply 8 by 4.

Conclusion

 25 19 20 13 14 15 7 8 9 10 1 2 3 4 5

6 step stairs.

 57 58 59 60 61 62 63 64 49 50 51 52 53 54 55 56 41 42 43 44 45 46 47 48 33 34 35 36 37 38 39 40 25 26 27 28 29 30 31 32 17 18 19 20 21 22 23 24 9 10 11 12 13 14 15 16 1 2 3 4 5 6 7 8
 50 42 43 34 35 36 26 27 28 29 18 19 20 21 22 10 11 12 13 14 15

I will start by taking a random 6 by 6 step stair. I can see that x = 10. If I write this out algebraically I would get: (x) (x + 1) (x+2) (x+3) (x+4) (x+5) (x+g) (x+g+1) (x+g+2) (x+g+3) (x+g+4) (x+2g) (x+2g+1) (x+2g+2) (x+2g+3) (x+3g) (x+3g+1) (x+3g+2) (x+4g) (x+4g+1) (x+5g)

I will now take a random 6 step stair from a 10 by 10 grid.

 95 85 86 75 76 77 65 66 67 68 55 56 57 58 59 45 46 47 48 49 50

If I wrote this out algebraically I find that it would be the same as in the 8 by 8 grid being: (x) (x + 1) (x+2) (x+3) (x+4) (x+5) (x+g) (x+g+1) (x+g+2) (x+g+3) (x+g+4) (x+2g) (x+2g+1) (x+2g+2) (x+2g+3) (x+3g) (x+3g+1) (x+3g+2) (x+4g) (x+4g+1) (x+5g).

If I simplify all this I would find that the equation would be 21x + 35g + 35. This represents the equation for a 6 step stair on any size grid.

I will now test the equation on a random 6 step stair from a 6 by 6 grid.

 31 25 26 19 20 21 13 14 15 16 7 8 9 10 11 1 2 3 4 5 6

x=1 and g=6. If I substitute these into the formula I will get (21 multiplied by 1) + (35 multiplied by 6) add 35. This will equal 266. If you add the numbers individually you will also get 266 which shows that the equation works.

Here is a table of my results

 Number of steps in staircase Equation 3 6x+4g+4 4 10x+10g+10 5 15x+20g+20 6 21x+35g+35

I will now try and find an equation for any size stair on any size grid. The final product will be an equation like this: (part1)x + (part2)g + (part3).

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

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# Related GCSE Number Stairs, Grids and Sequences essays

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