I have changed the grids to see whether or not there is a pattern. I kept the shape name as one all the time. I have done 5, 6,7and 8 grids.
There is clearly a pattern here; the totals are increasing by 4; so each time I increase the grid size by 1 the total increases by 4, there is a common difference.
From this information I predict that the 9th grid will have a total of 46.
Results:
My prediction was correct.
Algebraic Expression
I have drawn a 3-step stair with algebra on it, the g represents grid.
Totals:
(x)+(x+1) +(x+1+1) +(x+g) +(x+g+1) +(x+g+g) = 6x+4g+4
Comment:
The reason it’s 6x is because there are 6 stairs, each one with one x on it (x+x+x+x+x+x=6x), x represents the name of the shape.
The reason it’s 4g is because say for example you had a 5 grid, notice how every time you go up the total increases in accordance with the grid number; e.g.:
+5
So basically if it were any grid the number on top of one would always increase by the grid number. That is why the g is situated on top of the x in my algebra stairs; and since there are 4g’s we write it down in the formula as 4g.
The reason it’s +4 is simply because it’s (1+1+1+1). These ‘1’s’ are on my algebra stairs because in any stairs no matter what grid it is, you always go one across.
+1
So, my formula is: 6x+4g+4
I will now experiment it on 3 different grids to see if it works.
I will try it on a 3 grid, a 7 grid and a 10 grid.
My formula worked for all 3 grids, from this I can conclude that the formula, 6x+4g+4 will work on any grid provided that it’s a 3 step-stair.
The formula 6x+4g+4 is a linear formula. I have found a formula that will work on any 3-step stair, now I need to find a formula which will work on any sized step.
I will now change the stairs size to try and find a formula which will work on any stair size.
I will do a 3, 4, 5 and 6 step stairs on a 10 grid.
3-STEP STAIR
4-STEP STAIR
5-STEP STAIR
6-STEP STAIR
From this information I have found a link between triangle numbers and the numbers above:
Triangle numbers are; 1, 3, 6, 10, 15, 21, 28, 36, 45…
Notice how the stairs are made up of boxes or squares which are triangle numbers, e.g.:
In a 4-step stair there are 10 squares which make up the stairs.
In a 5-step stair there are 15 squares which make up the stairs.
Evidently then for each time you increase the step size the total amount of squares needed increases in accordance to the triangle numbers formula:
[n (n+1)]/2
So if we knew that the 3-step stairs had 6 squares and wanted to find out how many squares a 4-step stair had, we would do the following calculation:
[4(4+1)]/2= 10
This is the amount of squares in a 4-step stair.
I will now do the 3, 4, 5, 6 and 7 -step stairs in algebra.
3-step stairs
Formula= 6x+4g+4
4-step stairs
Formula= 10x+10g+10
5-step stairs
Formula= 15x+20g+20
6-step stairs
Formula= 21x+35g+35
7-step stairs
Formula=
28x+56g+56
I will now attempt to find a formula which will give a formula for any stair-size and any grid.
From the table above it is clear that the number before the ‘x’ is always a triangle number, therefore we know that the formula begins as follows:
[n (n+1)/2] x – this is the formula used to find ant triangle number.
From the table above we also learn that the last 2 numbers of the formula are always the same,
e.g. 4g+4 or 20g+20. So we know that if we find the formula for g then it will be the same for the last number.
Our series is:
1, 4, 10, 20, 35, 56 …- we need to find an equation for this.
Triangle series is the sum of natural numbers, which gives us:
1, 3, 6, 10, 15, 21, 28 … – the sum of triangle numbers gives you this:
1=1
1+3=4
1+3+6=10
1+3+6+10=20
1+3+6+10+15=35
1+3+6+10+15+21= 56
As you can see when you add the triangle numbers you get our series, so we know that it is the sum of triangle numbers.
Now we need to find a formula so we do the following:
1= 1 x ? =1
4= 2 x “ = 2 x 2
10= 3 x “ = 3 x 3⅓
20= 4 x “ = 4 x 5
35= 5 x “ = 5 x 7
56= 6 x “ = 6 x 9⅓
Some of the answers are not whole numbers, e.g. 3 x 3⅓, but are fractions with a denominator of 3. If we multiply everything by 3 we get:
1 x 3= 1 x 3
4 x 3= 2 x 6
10 x 3= 3 x 10 Notice how these are triangle numbers. However, it’s the value of n+1.
20 x 3= 4 x 15 So we substitute it with the triangle numbers formula except we use (n+1):
35 x 3= 5 x 21 [n (n+1) (n+1) +1)]/2, but remember we have to divide by 3 because we
multiplied by 3 earlier so the formula is: [n (n+1) (n+2)]/6
In our series the 4 is from a 3 step, the nth step would need to be (n-1) in
the above formula (Blue). We substitute n with (n-1):
[(n-1) ((n-1) +1) ((n-1) +2)]/6
We can simplify the formula into this:
[n (n+1) (n-1)]/6
So the universal formula is: [n (n+1)]/2 x + [n (n+1) (n-1)]/6 g+ [n (n+1) (n-1)]/6
n= step size
x= number of shape
g= grid number
I will now test to see if it works, I want to get the formula for a 3-step stair, so I do the following:
[3(3+1)]/2 x + [3(3+1)(3-1)]/6 g + [3(3+1) (3-1)]/6 which equals:
6x+4g+4 – My formula works for the 3-step stairs.
I will do the same, but this time for a 7 step stairs:
[7(7+1)]/2 x + [7(7+1) (7-1)]/6 g + [7(7+1) (7-1)]/6 which equals:
28x+56g+56
Clearly my formula works, through the use of algebra and my insight into the relationships of series and triangle numbers I was able to achieve this. I needed to use a variety of methods and techniques to complete the universal formula, always linking it back with previous formulas and tables.