The first difference between all of these is 60. We therefore know that the formula is to be a linear one. If we say that:
Total = a n + b
And test it using n = 5 and n = 15:
74 = 5a + b
134 = 15 a + b
We now have a simultaneous equation.
74 = 5a + b _
134 = 15a + b
60 = 10a
6 = a and b = 44
We can now put these numbers into the formula and find that
6n+44
is the formula to find the sum of any three step stair on a 10x10 grid.
To prove this I will put the algebraic values for each square in the three step stair.
We can now see that there are six squares in the stair (6n) and the numbers, which show how much they increase n by, add up to 44 (+44)
Step 2
Now we can begin to either vary the size of the grid or vary the size of the stair shape.
I will vary the size of the grid first because this is the easier option as there will still only be six numbers to add up in the three step stair.
Here are three grid examples with possible options for three step stairs highlighted.
I then did the same for a 6x6, 7x7 and 8x8 grid and came up with the table of results below by going through the same processes as in step one.
From this, by looking at the pattern, it is possible to determine the formula for gxg which is:
6n+4(g+1)
This can be proved by showing the algebraic values for each square in the stair.
There are 6 ns, 4 gs and the numbers add up to 4.
Now I will move on to changing the stair size. I will continue to use a 10x10 grid to keep a constant.
Here are some examples of a two step stair, a three step stair and a four step stair on a 10x10 grid.
Once again, by working out the algebraic values of each square within the stairs, as I have done below for two, three and four step stairs, one can find the formula for each step stair.
3n+11 6n+44 10n+110
I have also done this for five step stairs.
From this we can see a pattern. There is no same first difference, so the formula is not a linear one, there is no second difference so it is not a quadratic but the third difference is the same, 11, so therefore we know it is a cubic formula, looking something like this:
Total = as^3 + bs^2 + cs
where s = The number of steps on the stair
In order to work out this formula a simeltaneous equation must be used.
Firstly I will take the equations used for the 2 and 3 step stair.
3n + 11 6n + 44
These equations can then be manipulated and added to the formula, and then be worked out.
11 = 8a + 4b + 2c * 3
44 = 27a + 9b + 3c *2
33 = 24a + 12b + 6c
_
88 = 54a + 18b + 6c
55 = 30a + 6b
55/30 = 1 5/6
From here we can then work out the value of c
11 = 8a + 4b + 2c * 27
44 = 27a + 9b +2c *8
297= 216a + 108b + 54c
-
352 = 216a + 72b + 24c
-55 = 36b + 30c
-55/30 = -1 5/6
To test to see if the formula works I will first convert the fractions to top heavy ones.
1 5/6 = 11/6 -1 5/6 = - 11/6
Now the fractions will be placed into the formula.
(11/6*3^3) + (- 11/6*3) = 44.
