4-step stairs:
By adding up all the Xs all the gs and all the numbers I got this:
X+X+1+X+2+X+3+X+g+X+g+1+X+g+2+X+2g+X+2g+1+X+3g=10X+10g+10 This is the formula for any four-step stairs on any size grid.
To prove that this formula worked I drew two grids of difference sizes and calculated the total of all the numbers in the 4-step stair.
By using the formula 10X+10g+10=S, I worked out the total of the numbers inside the red area of the 4-step stair.
(10*36)+(10*10)+10=260+100+10= 470
Then I added up all the numbers in the 4-step stair to see if it gave the same answer.
36+37+38+39+46+47+48+56+57+66= 470.
By using the formula 10X+10g+10=S, I worked out the total of the numbers inside the blue area of the 4-step stair.
(10*18)+(10*7)+10= 180 + 70+ 10= 260
Then added up the numbers in the 4-step stair as I did before.
18+19+20+21+25+26+27+32+33+39= 260
The two examples above prove that the formula 10x+10g+10 calculates the total of the numbers inside the area covered by a 4-step stair on any grid size.
5 step stairs:
By adding all the Xs, all the gs and all the numbers together I got:
X+X+1+X+2+X+3+X+4+X+g+X+g+1+X+g+2+X+g+3+X+2g+X+2g+1+X+2g+2+X+ 3g+X+3g+1+X+4g = 15X+20g+20. This is the formula for all 5-step stairs on any size grid.
To prove this formula works for all size grids and therefore works in general I drew two different sized grids and did the following calculations:
By using the formula 15X+20g+20=S, I worked out the total of the numbers inside the green area of the 5-step stair.
(15*25)+(20*10)+20=595
Then I added all the numbers in the 5-step stair to see if it came up with the same answer:
25+26+27+28+29+35+36+37+38+45+46+47+55+56+65=595.
By using the formula 15X+20g+20=S, I worked out the total of the numbers inside the turquoise area of the 5-step stair:
(15*1)+(20*5)+20=15+100+20=135
I then added up all the numbers in the 5-step stair to see if it gave the same number:
1+2+3+4+5+6+7+8+9+11+12+13+16+17+21=135
These two diagrams prove that 15X+20g+20=S on all sized grids.
6-step stairs:
By adding all the Xs all the gs and all the numbers up together I got this:
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+2g+3+X+3g+X+3g+1+X+3g+2+X+4g+X+4g+1+X+5g = 21X+35g+35.
To prove that this formula works I drew up two grids and used the formula to calculate the total of the numbers inside the 6-step stair and saw if it was the right answer on both grids.
I did this calculation to see if the formula works.
21X+35g+35 = (21*11) + (35*8) + 35 = 546
11+12+13+14+15+16+19+20+21+22+23+27+28+29+30+35+36+37+43+44+51= 546
As above I did this calculation to see if the formula works.
21X+35g+35 = (21*1) + (35*6) + 35 = 21+210+35 = 266
1+2+3+4+5+6+7+8+9+10+11+13+14+15+16+19+20+21+25+26+31= 266
This proves that the formula works on any sized grid. Therefore, 21X+35g+35 = S is the formula for 6-step stairs on any grid size.
After researching and finding the formulas for 4,5 and 6 step stairs as well as 3-step stairs I took the results and put them in a table to look for patterns.
n X g +
3 6X + 4g +4
4 10X +10g +10
5 15X +20g +20
6 21X +35g +35
I soon realised that all the numbers in column X (that are in red), are all triangle numbers. This made sense because the layout of a triangle number looks like this:
.
. .
. . .
and the layout for a step-stair looks like this:
The triangle number above = 6. This is because the 1 on the top row+ 2 on the 2nd row and 3 on the 3rd row = 6, 1+2+3=6.
If you add up all the Xs in the three-step stair above it = 6X. So this is how I know that all the Xs in a step-stair will always be a triangle number, because all step-stairs have the same layout = X on the top row. 2X on the second row if you add them both together, 3X on the 3rd row in you add them all together, 4X on the 4th row if you add them all together, this means that it is always X+2X+3X+4X etc in terms of the numbers of X per row, which will always equal a triangle number when you add them all up, this is explained in the diagram below this paragraph. This works for whatever the step size is.
X+2X+3X=6X
I worked out that in all step stairs the first part of the equation starts with the triangle number of n, the step stair size then you times this by X. So when n=3, the third triangle number= T3 = 6 so it is 6X. This works because, as you can see in the table above, the value of X when n, the stair size number = 3, is 6X.
All triangle numbers are worked out like this, for instance:
1+2+3+4+5=15 this is the 5th triangle number as there is 5 numbers that make up the total figure.
Other triangle numbers:
1=1
1+2=3
1+2+3=6
1+2+3+4=10
1+2+3+4+5=15
1+2+3+4+5+6=21
1+2+3+4+5+6+7=28
Etc.
In my formulas I used the notations for the following numbers.
1 = T1
3 = T2
6 = T3
10 = T4
15 = T5
21 = T6
28 = T7
Etc
Tn = the triangle number followed by which triangle number it is, e.g. T1 = the first triangle number = 1. T2 = the second triangle number = 3.
I then decided to try and work out the formula for the nth triangle number, so that I could write the first part of the formula for any size step stair on any sized grid.
I constructed a table comparing n to the triangle numbers.
Triangle Numbers
n 1st difference
- 1
2
2 3
3
3 6
4
- 10
5
5 15
6
6 21
I first of all wrote down the things that I noticed that stayed the same all the way through:
n always = n
The difference between the triangle numbers always = n+1.
With the knowledge that the things above always remain constant I can multiply them together and see what answer they give.
So,
n(n+1) = 4(4+1)= 20. The fourth triangle number is exactly ½ that total. So,
n(n+1) = Tn, this is the formula that I think is the formula for any triangle number.
2
I can prove this by saying n=5. This should give the 5th triangle number.
n(n+1) = 5(5+1) = 15. The 5th triangle number is 15. This and the working
2 2
shown above proves that the formula works.
This proven, I can now create the 1st part of the formula for any step stair on any grid size. I make n the step-stair size.
So the first part of the formula so far looks like this:
n(n+1) X + (blank)g+ (blank)
2
The n(n+1) X, bit is prove by this:
2
n, the step-stair size = 3, 3(3+1) X = 6X.
2
As you can see in my results on past pages show that the number of Xs in a 3 step-stair = 6. This proves that this starting formula works.
The second thing I realised was that the bits below coloured red always stayed the same, and equal in value. This meant that the formula for both of them would be the same, only, the one that describes g will have g written after the brackets of the formula for parts of the equation labelled red:
6X+4g+4
10X+10g+10
15X+20g+20
21X+35g+35
So the formula so far looks like this:
n(n+1) + (blank) g + (blank) How I solved the blanks is written below.
2
I investigated the step stair patterns and found another really helpful pattern. This is described below:
The numbers outside the grid on the right hand side of each row are the total of all the numbers within the row itself, these are written in red. As you can see the numbers totalled up in each row equals a triangle number. The triangle number is equal to the number of blocks there is in its row, e.g. in the first row there is one block this means that the triangle number is T1. This will always remain constant because every row has the numbers that are added up starting at one and then going 1+2+3+4+5. This kind of equation will always give a triangle number and seeing as this remains the same it means that there will always be a triangle number at the total for each row. I then realised that by adding T1 to T4 together it gives the total 20. This is the number that occurs in the formula for a 5-step stair in the form of 20g+20. The step stair above is a 5-step stair. This means by counting up all the numbers per row and adding all the triangle numbers minus the last in the series of triangle numbers it will give you the answer to the last two parts of the equation for any step stair on any sized grid. This is the (blank)g +(blank) bit. So I needed to use a formula to add up a range of numbers that started at 1 and ended at n. I needed to do this to fill in the formula gaps. There was one formula that I already knew that added up a range of numbers in this way. It was this:
n
Σ r =
r =1
I this formula the sigma means “the sum of”. So the formula means the sum of r (the number you start at) to n ( the number you finish at). The r = bit means the sum of all the numbers between r and n when placed in ascending order. This formula will only add up numbers that are in series with each other e.g. 1,2,3,4,5… etc.
So if I put 5 in the place of n;
5
Σ r = 1+2+3+4+5 = 15
r = 1
I can now change this formula to become the formula that will be put in my final equation for any step stair on any size grid. If I place Tr = instead of the r = then it will mean all the triangle numbers added up from r = 1, so the first triangle number to whatever triangle number it is on the top of the sigma. So, if the number at the top of the sigma is 4 then it will be all the triangle numbers added up from T1 to T4 inclusive. In the pattern I noticed above I realised that if you added up all the triangle numbers that are made by adding up then numbers in its row, minus the last triangle number it gives the value of the (blank)g +(blank) bit in every step stair. So if I put n-1 at the top of the sigma it will mean that the formula will add up all the triangle numbers from T1 to Tn – 1. If I didn’t do this then the total, which I hope will fill the blanks for the last two parts of the final equation, will always be too high by the amount of the highest triangle number in the series of the triangle numbers that I added up. The n in this equation will be the step stair size. So if the step stair is a 3-step stair then the value at the top will read (3-1) = 2. This will mean that I the sigma formula will add up all the triangle numbers from the first triangle number to the second triangle number. I will keep r = 1 the same because I want to add up all the triangle numbers starting from the first triangle number. This means that the formula now looks like this:
(n-1)
Σ Tr =
r = 1
So if I put 3 in place of n, the step-stair size, I get this equation.
(3-1) = 2
Σ Tr = T1 + T2 = 1+3 = 4
r = 1
So, for a 3 step stair the value of (blank)g + (blank) = 4. If you look up the formula for a 3-step stair is 6X+4g+4. I then proposed that the formula:
(n-1) (n-1)
n(n+1) + Σ Tr = g + Σ Tr =
2 r = 1 r = 1
Is the formula for any step stair on any sized grid.
I tried it on the 4 step stairs I had investigated before. The 3-step stair, the 4-step stair, the 5-step stair and the 6-step stair.
So, by replacing n with 3, for a 3-step stair I get this:
(3-1) (3-1)
3(3+1) + Σ Τr = T1 + T2 g + Σ Tr = T1+T2
2 r =1 r = 1
So by using this formula that I have explained above:
The formula gives: 6X+4g+4. T1 + T2 = 4, (1+3).
This is the formula for a 3-step stair on any size grid. The formula works.
4-step stair:
(4-1) (4-1)
4(4+1) + Σ Τr = T1 + T2 + T3 g + Σ Tr = T1+T2 + T3
2 r = 1 r = 1
So by using this formula that I have exlained above:
The formula gives: 10X+10g+10. T1 + T2 + T3 = 10 (1+3+6)
This is the formula for a 4-step stair on any size grid. The formula works.
5-step stair:
(5-1) (5-1)
5(5+1) + Σ Τr = T1 + T2 + T3 + T4 g + Σ Tr = T1+T2 + T3 + T4
2 r = 1 r = 1
By using the formula I have explained above;
The formula gives: 15X+20g+20. T1 + T2 + T3 + T4 = 20, (1+3+6+10)
This is the formula for a 5-step stair on any sized grid. The formula works.
6-step stair:
(6-1) (6-1)
6(6+1) + Σ Τr = T1 + T2 + T3 + T4 + T5 g + Σ Tr = T1+T2 + T3 + T4 + T5
2 r = 1 r = 1
By using the formula I have explained above:
The formula gives: 21X+35g+35. This is the formula for a 6-step stair on any sized grid. The formula works.
I have now proved that the formula I found works for any step stair size on any size grid.
(n-1) (n-1)
n(n+1) + Σ Tr = g + Σ Tr =
2 r = 1 r = 1
This concludes my first coursework submission. Submitted on:
8th December 2003.
William Murray