= 120 – 10
= 110
Formulae: 10n + 110 = Total
Testing the formula:
Total = 10n + 110
Total = 10 * 3 + 110
Total = 30 + 110
Total = 140
5 Step Stairs:
Difference = +15 +15
+15 +15
+15
As we can see, the stair total for a 4-step stair using 1 as the base number on a 10 by 10-number grid is 235 as 1 + 2 + 3 + 4 + 5 + 11 + 12 + 13 + 14 + 21 + 22 + 23 + 31 + 32 + 41 = 235. I can therefore conclude that the general equation for a 5-step stair on a 10 by 10 number grid is 15n + 220 = stair total where n is the base number. (Shaded box).
Total of step containing 1 as its base number – Difference
= 235 – 15
= 220
Formulae: 15n + 220 = Total
Testing the formula:
Total = 15n + 220
Total = 15 * 5 + 220
Total = 75 + 220
Total = 295
6 Step Stairs:
Difference = +21 +21
+21
As we can see, the stair total for a 6-step stair using 1 as the base number on a 10 by 10-number grid is 406 as 1 + 2 + 3 + 4 + 5 + 6 + 11 + 12 + 13 + 14 + 15 + 21 + 22 + 23 + 24 + 31 + 32 + 33 + 41 + 42 + 51 = 406. I can therefore conclude that the general equation for a 6-step stair on a 10 by 10 number grid is 21n + 385 = stair total where n is the base number. (Shaded box).
Total of step containing 1 as its base number – Difference
= 406 – 21
= 385
Formulae: 21n + 385 = Total
Testing the formula:
Total = 21n + 385
Total = 21 * 4 + 385
Total = 84 + 385
Total = 469
7 Step Stairs:
Difference = +28
+28
+28
+28
As we can see, the stair total for a 7-step stair using 1 as the base number on a 10 by 10-number grid is 644 as 1 + 2 + 3 + 4 + 5 + 6 + 7 + 11 + 12 + 13 + 14 + 15 + 16 + 21 + 22 + 23 + 24 + 25 + 31 + 32 + 33 + 34 + 41 + 42 + 43 + 51 + 52 + 61 = 644. I can therefore conclude that the general equation for a 7-step stair on a 10 by 10 number grid is 28n + 616 = stair total where n is the base number. (Shaded box).
Total of step containing 1 as its base number – Difference
= 644 – 28
= 616
Formulae: 28n + 616 = Total
Testing the formula:
Total = 28n + 616
Total = 28 * 3 + 616
Total = 84 + 616
Total = 700
8 Step Stairs:
Difference = +36
+36
+36
As we can see, the stair total for a 8-step stair using 1 as the base number on a 10 by 10-number grid is 960 as 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 21 + 22 + 23 + 24 + 25 + 26 + 31 + 32 + 33 + 34 + 35 + 41 + 42 + 43 + 44 + 51 + 52 + 53 + 61 + 62 + 71 = 960.
I can therefore conclude that the general equation for an 8-step stair on a 10 by 10 number grid is
36n + 924 = stair total where n is the base number. (Shaded box).
Total of step containing 1 as its base number – Difference
= 960 – 36
= 924
Formulae: 36n + 924 = Total
Testing the formula:
Total = 36n + 924
Total = 36 * 2 + 924
Total = 72 + 924
Total = 996
Formulas:
Final Formula:
Section 1:
= First Difference
= Second Difference
The data above is summarised in the table below.
Observation:
The differences between the nth terms were increasing which meant that I had to use quadratic sequences. The general term of a quadratic sequence is:
un = AN 2 + BN + C
a= the second difference in the sequence over 2.
b = is the coefficient of the term in n 2.
c = is the numerical value or constant value.
a = 2nd difference / 2
a = 1 / 2
Counting back, the zeros term is 0. So when n = 0, un = 0. When ‘n’ = 0, both ‘n 2’ and ‘bn’ are equal to 0 so ‘c’ = 0.
C = 0
To find the value of ‘b’, we look at the first term. When ‘n’ = 1, the value of u is 2.
un = n + bn
= 1 + b
2 = 1 + b
b = 1 / 2
1/2 s2 + 1/2 s
Section 2:
= First Difference
= Second Difference
= Third Difference
The data above is summarized in the table below.
Nth term = an 3 + bn 2 + cn + d
When n = 1 13a+12b+1c+d=6 Equation 1: a + b + c + d = 0
When n = 2 23a+22b+2c+d=20 Equation 2: 8a + 4b + 2c + d = 11
When n = 3 33a+32b+3c+d=52 Equation 3: 27a + 9b + 3c + d = 44
When n = 4 43a+42b+4c+d=108 Equation 4: 64a + 16b + 4c + d = 110
Subtracting equation 1 from each of the other 3 gives:
Equation 5 7a+3b+c=11
Equation 6 26a+8b+2c=44
Equation 7 63a+15b+3c=110
Subtracting 2 times equations 5 from equation 6 gives:
Equation 8 12a+2b=22
Subtracting 3 times equations 5 from equation 7 gives:
Equation 9 42a+6b=77
And subtracting 3 times equation 8 from equation 9 to work out letter a.
To work out b we can now substitute it into equation 8 or 9. Form the values gained, we then use equation 5, 6 or 7 to find c and then these values into one of the original equations to find d.
Equation 5:
8a + 4b + 2c + d = 11
- a + b + c + d = 0
7a+3b+c = 11
Equation 6:
27a + 9b + 3c + d = 44
- a + b + c + d = 0
26a+8b+2c=44
Equation 7
=64a + 16b + 4c + d = 110
- a + b + c + d = 0
= 63a+15b+3c=110
Equation 8:
7a+3b+c=11 * 2
14a + 6b + 2c = 22
26a + 8b + 2c
- 14a + 6b + 2c = 22
12a + 2b = 22
Equation 9:
7a+3b+c=11 * 3
21a + 9b + 3c = 33
63a + 15b + 3c = 110
- 21a + 9b + 3c = 33
42a + 6b = 77
Working out letter a:
12a + 2b = 22 * 3
36a + 6b = 66
42a + 6b = 77
- 36a + 6b = 66
6a = 11
a = 11 / 6
Working out letter b:
12a + 2b = 22
12 * 11 / 6 + 2b = 22
22 + 2b = 22
2b = 22 - 22
b = 0 / 2
b = 0
Working out letter c:
26a + 8b + 2c = 44
26a * 11 / 6 + 8 * 0 + 2c = 44
47.66666667 + 2c = 44
2c = 44 – 44.66666667
2c = - 3.66666667
c = - 3.66666667 / 2
c = - 1.833333335
Working out letter d:
a + b + c + d = 0
11 / 6 + 0 + -1.833333335 +d = 0
-0.000000001 + d = 0
d = 0 - -0.000000001
d = - 0.000000001
Letter a, b, c, and d stands for:
S = Number of stairs
N = Base Number
Final Formula:
[( 1 / 2 s 2 + 1 / 2 s) n] + (as 3 + bs 2 + cs + d )
Testing the formula:
The formula can be tested and verified by calculating a 5-step stair with the base number equalling to 1.
[( 1 / 2 s 2 + 1 / 2 s) n] + (as 3 + bs 2 + cs + d )
= [(1 / 2 * 5 2 + 1 / 2 * 5) n] + (11 / 6 * 5 3 + 0 * 5 2 + - 1.833333335 * 5 + 0.000000001)
= [(12.5 + 2.5) n] + 11 / 6 *125 + 0 * 25 + -9.166666675 + 0.000000001)
= 15n + (229.1666667 + 0 + -9.166666674)
= 15n + (229.1666667 + -9.166666674)
= 15n + 220
15n + 220
= 15 * 1 + 220
= 15 + 220
= 235
The answer establishes the formula to be correct.
We can now introduce algebra in order to find a common formula to find the stair total for a 3-step stair on a 10 by 10-number grid.
If this stair is stair ‘x’ then the numbers in the stair will be
(x) + (x+2) + (x+3) + (x+10) + (x+11) + (x+20) this can also be written as
6x + 44 = stair total.
Using stair 7 we are now going to use my formula to find out its stair total and see if my formula is correct. x is equal to 7 as 7 is the number in the bottom left hand corner of stair7 and x is the number in the bottom left hand corner of stair x. .
6x + 44 = stair total
(6 x 7) + 44 = 86
So my formula says that the stair total of this stair is 86.
Now we can test this.
7 + 8 + 9 + 17 + 18 + 27 = 86
This is true. Therefore we can now say that the general formula for any 3-step stair on a 10 by 10-number grid is 6x + 44 = stair total. .
We are now going to investigate further the relationship between stair totals and other step stairs on other number grids.
The numbers inside the stair and the formula for finding the stair total varies depending on the grid size:
For a 10 by 10 Grid:!
But for a 9 by 9 Grid:
We can see that the formula for a 9 by 9 grid is 6x + 40 = stair total. This is 4 less than the formula for the 10 by 10 grid.
In order to prove this I will use stair 20 on a 9 by 9 grid.
6x + 40 = stair total
(6 x 20) + 40 = 160.
20 + 21 + 22 + 29 + 30 + 38 = 160
I can therefore conclude that the general equation for a 3-step stair on a 9 by 9 number grid is 6x + 40 = stair total where x is the number in the bottom left hand corner of the stair shape.
We can now say that in order to find the equation for an 11 by 11-number grid all we have to do is add 4 to the equation of a 10 by 10 grid. If we do this then the equation will be 6x + 48 = stair total.
I am now going to check this equation with stair 12 on an 11 by 11 number grid.
6x + 48 = stair total
(6 x 12) + 48 = 120
12 + 13 + 14 + 23 + 24 + 34 = 120 o9xuE
This is true so I conclude that the stair total for any 3-step stair on an 11 by 11 number grid is 6x + 48= stair total.
We can now say that every time you increase the size of the grid by one (going from a 9by9 number grid to a 10by10 number grid) you have to add four more to the equation. Therefore if you decrease the size of the grid by one you must also decrease the equation by 4.
With this knowledge I can now find a general equation for any 3-step stair on any size grid.
Grid Size Formula:
7 by 7 = 6x + 32
8 by 8 = 6x + 36
9 by 9 = 6x + 40
10 by 10 = 6x + 44
11 by 11 = 6x + 48
So from the results table I can see a linear pattern, which confirms my theory. Every time you increase the grid size by 1 you increase the formula by 4
Here is the diagram that illustrates the equation for any 3-step stair on any size grid:
g = grid size
x = stair number
x + x + x + x + x + x = 6x
g + g + 2g = 4g
1 + 1 + 2 = 4
Stair total = 6x + 4g + 4
In order to prove that my formula is correct I am going to test it by using it to find the stair total for stair 17 on a 10 by 10 size grid.
x = 17
g = 10
6x + 4g + 4 = stair total
6 (17) + 4 (10) + 4 = 146
17 + 18 + 19 + 27 + 28 + 37 = 146
This is correct. Therefore I conclude that the general equation for finding a 3-step stair on any sized grid is 6x + 4g + 4 = stair total.
I have 6x as there are always 6x in any 3-step stair.
I add 4g as this is the number found when we increase the grid size by one.
Finally I add 4 to these two numbers because 1 + 1 + 2 equals 4.
Also every time you move one square to the right you increase the stair total by 1 and every time you move one square down up you increase the total by g.
Changing the Size Of The Stair:
2-Step Stair
3x +1+g = stair total is the equation for any 2-step stair on any size grid.
3-Step Stair
Stair total = 6x + 4g + 4 Is the equation for any 3-step stair on any size grid.
4-Step Stair
Stair total = 10x + 10g + 10 is the equation for any 4-step stair on any size grid.
5-Step Stair
The equation for any 5-step stair on any size grid is:
15x + 20g + 20 = stair total.
6-Step Stair
21x + 35g + 35 = stair total is therefore the equation for a 6-step stair on any size grid.
Now that I have found these equations I can see three sets of sequences forming that will help to form the final equation for any size stair on any size grid.
It is clear that these three sets of sequences are the co-efficient in front of x sequence, the coefficient in front of g sequence and the constant sequence. I am now going to investigate each separate sequence in order to find the general equation for the coefficient of x sequence, the general equation for the co-efficient of g sequence and the general equation for the constant sequence. The final product will be an equation in the following form :
(Section 1) x + (Section2) g + (Section3)
Section 1: The Coefficient In Front Of x Sequence
As this sequence has a constant number at the 2nd difference we can easily say that this is a quadratic sequence. Therefore the first part of the equation is going to be :.
2nd difference / 2 n2 = 1/2 n2
We can now use this value with different values of n: .
n= 1 , 2 , 3 , 4 , 5f.
1/2 n2 = 0.5 , 2 , 4.5 , 8 , 12.5 7
Now that we know the values of n for 1/2 n2 all we have to do is subtract it from the original values of n in the quadratic sequence.
Original Sequence = 3, 6, 10, 15, 21
1/2 n2 =
0.5 , 2 , 4.5 , 8 , 12.5
2.5 , 4 , 5.5 , 7 , 8.5
1.5 , 1.5 , 1.5 , 1.5
We can now conclude that the co-efficient in front of n will be 1.5n or 3/2 n.
Since we now know now the beginning of this sequence formula to be 1/2 n2 + 3/2n we can now find the constant for this sequence.
Original Sequence = 3 , 6 , 10 , 15 , 21
1/2 n2 + 3 / 2 n = 2 , 5 , 9 , 14 , 20
Result = 1 , 1 , 1 , 1 , 1
Looking at these results we can therefore say that the constant for this sequence is 1.
We can now say that the nth term for this quadratic sequence is :
1/2 n2 + 3/2n + 1 = co-efficient of x.
Section 2: The Co-Efficient In Front of g Sequence:
The coefficient in front of g has the following sequence:.
Because this sequence has a constant number at the 3rd difference it is easy to see that it is a cubic sequence. If this sequence is cubic it will follow this formula:
an3 + bn2 + cn + d:
If n = 1 , 2 , 3 , 4 , 5
a+b+c+d , 8a+4b+2c+d , 27a+9b+3c+d , 64a+164+4c+d , 125a+25b+5c+d , 7a+3b+c , 9a+5b+c , 37a+7b+c , 61a+9b+c, 12a+2b , 18a+2b , 24a+2b
6a , 6a
I am now going to match the numbers in the cubic sequence formula above to the numbers that correspond to their positions in the co-efficient in front of g sequence. 6a =
a = 1/6
12a+2b = 3
12(1/6) + 2b = 3
2 + 2b = 3
2b = 1
b = 0.5
7a+3b+c= 3
7(1/6)+3(1/2)+c = 3
c= 1/3
a+ b + c + d = 1
1/6 + ½ + 21/3 + d = 1.
d = 0
Therefore the equation for this part of the sequence will be :
(1/6 n3 + 1/2n2 + 1/3n) = coefficient of g
Section 3: The Sequence For The Constant
The sequence for the constant will go exactly the same as the sequence for the co-efficient of g as all these values are the same. Therefore the equation for this part of the sequence will be :
(1/6 n3 + 1/2n2 + 1/3n) = the constant
Conclusion
Now that we have finished the three parts of the sequence we can put them together in order to make one equation for any size step-stair on any size grid.
(1/2 n2 + 3/2n + 1) x + (1/6 n3 + 1/2n2 + 1/3 n) g + (1/6 n3 + 1/2n2 + 1/3 n) = stair total
x means the number of squares in the stair shape..
g stands for grid size.
n is the equation number. (If you don't know the equation number it is the same number as the number of squares in the 2nd column from the left of the stair shape).
Reference:
-
Internet:
- Higher GCSE for Edexcel
- Edexcel GCSE Mathematics higher course
- GCSE Mathematics Revision Guide Higher Level
