I will be using the same method but in algebra form.
χ
χ+χ+1+χ+10 = 3χ + 11
χ+χ+1+χ+2+χ+10+χ+11+χ+20= 6χ+44
χ+χ+1+χ+2+χ+3+χ+10+χ+11+χ+12+χ+20+χ+21+χ+30 =10χ+110
15χ+1+2+3+4+10+11+12+13+20+21+22+30+31+40=
15χ+220
Now that I have found the formula for the step stairs shown above, I will be taking it further in order to get an overall formula for any size step stair which works on a 10 by 10 grid.
This is how I will be trying to find an overall formula, by working it out step by step; the way in which I will be trying to find out this formula is by laying out like this, as shown below:
Differences of the coefficient of χ
Step stairs 2nd 1st Formula
1 χ
2
2 1 3χ +11
3
3 1 6χ +44
4
4 1 10χ +110
5
5 1 15χ +220
6
6 21χ +385
At first I worked out the differences of χ. By looking at this table, I have noticed that this is a quadratic sequence because the 2nd difference is the same. The formula for a quadratic is an²+ bn+c
To find out c:
n 1 2 3 4 5 6
c= (0) 1 3 6 10 15 21
1 2 3 4 5 6
2nd difference 1 1 1 1 1
To find out a, you divide the 2nd difference by 2. Therefore a = ½
a= ½ c= 0
So far the formula is ½ n² + bn
To work out b, I am going to write down ½ n² and see what times n gives me the coefficients of χ.
½ n² ½ 2 4.5 8 12.5 18
n 1 2 3 4 5 6
C= (0) 1 3 6 10 15 21
1 2 3 4 5 6
2nd difference 1 1 1 1 1
b= ½
The formula is ½ n² + ½ n
This formula is the co-efficient of χ
Now I need to find out a formula for numbers you add on.
Step stair Formula
1 χ +0 (11x0)
1
2 3χ+11 +11 (11x1)
3
3 6χ +44 +44 (11x4)
6
4 10χ+110 +110 (11x10)
10
5 15χ+220 +220 (11x20)
15
6 21χ+385 +385 (11x35)
(I am labelling these number as M)
By looking at the number you add on I have found out that these numbers are multiples of 11.
The difference of M is exactly the same as the co- efficient of χ. so the formula for the differences for M is: ½ n² + ½ n. I think that the formula for M is linked with the differences of M.
I believe this formula that I am trying to work out will be a cubic because the 3rd difference is the same.
Looking at the first line, in order to get 0, I need to multiply 1 (½ n² + ½ n) by 0. In this case n – 1= 0, so (n-1) (½ n² + ½ n) = 0.
Now I will be doing the same thing on the second line to get 1.
(n-1) (½ n² + ½ n) = (2 -1) x 3= 3, which doesn’t equal 1
Third line: (n-1) (½ n² + ½ n) = (3-1) x 6 = 12, which does not equal 4
Now if I divide the answers by 3, it gives me the answer I want.
3/3=1
12/3=4
Trial for 4th line: 1/3 (n-1) (½ n² + ½ n) = 11/3 (4-1) x 10 = 10
Formula: 1/3 (n-1) (½ n² + ½ n)
This is the formula for M, to get the number you add on you multiply by 11.
11/3 (n-1) (1/²n²+1/²n).
Therefore to work out any n step stair total is T= (1/2n² +1/2n) χ + 11/3 (n-1) (1/2n²+1/2n)
I am now going to test if my formula works:
(½ n² + ½ n) χ+ 11/3 (n-1) (1/2 n²+1/2 n)
(1/2 16 + ½ 4) χ + 11/3 (4-1) (1/2 16 + ½ 4)
10χ + 11/3 x 3 x 10
=10 χ + 110
This was to find out the 4 step stair.
Now to check if it works for 8 step stair
(1/2 8 + 4) χ+ 11/3 (8-1) (½ 8 + ½ 8)
(32 + 4) χ + 11/3 (7) (32+44)
36X 11/3 χ 7 X 36
36X + 924 Therefore I have found out a formula for 8 step stair
Now I am going to find out a formula for any size step on any size grid.
χ +χ+1+χ+2+χ+3+χ+A+χ+A+1+χ+A+2+χ+2A+χ+2A+
χ+3A.
Step stair Formula Difference
n=1 χ +0
1
n=2 3χ+A+1
3
n=3 6χ+4A+4
6
n=4 10χ+10A+10
10
n=5 15χ+20A+20
15
n=6 21χ+35A+35
Therefore by looking at the differences I can see that it is the same as the co=efficient of χ. Therefore the formula for any size grid and step stair is:
T= (1/2 n² + ½ n) χ+ 1/3 (n-1) (1/2 n² + 1/2n) a + 1/3 (n-1) (1/n² + 1/2n)
This is the formula for the grid. However I believe that you can simplify this formula which now looks like this:
T= (1/2 n² + ½ n) χ+ 1/3 (n-1) a + 1/3 (n-1)
You can simplify this formula even further:
T= ½ (n² + n) (χ +1/3 (n-1) (a+1))
In conclusion I state that the formula that I have found works for the step stair, and grid. Therefore it is useful to have theses formulas because it makes problem solve in an efficient way, therefore it allows the problem to be solved in the quickest way.