I will begin by using the same method as I used for finding the formula for 3 step stairs by taking all the 4 step stairs on the bottom row and finding the sum.
1)1+2+3+4+11+12+13+21+22+31=120
2)2+3+4+5+12+13+14+22+23+32=130
3)3+4+5+6+13+14+15+23+24+33=140
4)4+5+6+7+14+15+16+24+25+34=150
5)5+6+7+8+15+16+17+25+26+35=160
6)6+7+8+9+16+17+18+26+27+36=170
7)7+8+9+10+17+18+19+27+28+37=180
Using the formula Un=ab+c I can tell that the terms go up in tens. This tells me that the nth term will include 10 lots of n or 10n. For the first term n=1, so 6n=6. But the first term is 120 which is 110 more than 10n. This suggests that the formula is 10N+110.
I will now test the formula on a random 4 step stair.
Step 1) Locate n which is in bottom left corner.
N=33
Step 2) Insert n into formula 10n+110
Step3) (33 multiplied by 10) +110= 540 which equals the sum of the numbers.
The formula has worked and with this I can identify the sum of any 4 step stair.
Algebraically proving
Another way to prove my formula is to use algebra.
I will now use this method to obtain the formula for 5 and 6 step stairs as it is simple and easier to use.
5 step stairs
If we take a random 5 step stair:
I can see that 45 is n so therefore if I replace it with x I will have:
(X) + (X+1) + (X+2) + (X+3) + (X+4) + (X+10) + (X+11) + (X+12) + (X+13) + (X+20) + (X+21) + (X+22) + (X+30) +(X+31) + (X+40)
If I add these together I get 15X + 220. Then I substitute n back in and the formula is 15n +220.
I will now test my formula by picking a random 5 step stair.
6 step stairs.
If I take a random 6 step stair:
I can see that the number two is n. Therefore if I replace it with x I will get:
(X) (X+1) (X+2) (X+3) (X+4) (X+5) (X+10) (X+11) (X+12) (X+13) (X+14) (X+20) (X+21) (X+22) (X+23) (X+30) (X+31) (X+32) (X+40) (X+41) (X+50)
If I add these together I will get 21x+385. Then I substitute n back into the formula which gives me 21n+385.
I will now test my formula by selecting a random 6 step stair.
I have now successfully found formulas for 3, 4, 5 and 6 step stairs.
From this table I have noticed that the numbers before n are triangular. Therefore I assume that the formula for a seven step stair will include 28n.
I will now test this on a seven step stair to see if this is true.
Also I have noticed that the number of stairs corresponds to the first term of the formula.
To get from 3 to 6 you multiply by 2.
To get from 4 to 10 you multiply by 2.5.
To get from 5 to 15 you multiply by 3.
To get from 6 to 21 you multiply by 3.5.
If I put this in a table:
The difference between the terms is ½ so my formula must start with ½ n. If I take term 4 I will half it to get two then I will add half to get to 2.5. This applies to all the terms. Therefore my formula is ½ n+ 0.5. This shows me what to multiply the number of step stairs by to find out the first term of the formula.
I will take an 8 step stair as an example to prove my formula.
If I put 8 in the formula it would be ½ of 8 which is 4. Then add 0.5 which is 4.5. This shows me that I have to multiply 8 by 4.5 to find the first term of the formula which would be 36. Therefore the equation will be n(½ n+0.5).
With this equation I can now identify the first term of the equation for any size of step stair.
Different size grids
First of all I am going to collect all the 3 step stairs on the bottom row.
1+2+3+9+10+17=42
2+3+4+10+11+18=48
3+4+5+11+12+19=54
4+5+6+12+13+20=60
5+6+7+13+14+21=66
6+7+8+14+15+22=72
I will take a random 3 step stair to help prove my formula is correct.
To find the formula for any 3 step stair I must algebraically show the formula.
(x) (x+1) (x+2) (x+g) (x+g+1) (x+2g) = 4g + 6x +4 = 8 stair step.
(x) (x+1) (x+2) (x+g) (x+g+1) and (x+2g) = 4g+6x+4= 10 stair step
G represents size of grid. (If it was eight by eight grid then g would be 8, if it was ten by ten then g would be 10.)
Both the formulas are the same which means that they apply to 8 by 8 and 10 by 10 grids. I will now test this rule on a random 3 step stair on a 6 by 6 grid.
Therefore I can conclude that the general equation for finding a 3 step stair on any sized grid is 6x + 4g + 4 = stair total.
4 step stair on different size grids
I will now take a 4 step stair at random.
I can see that nine is x so therefore the other numbers can be written as:
(x) (x+1) (x+2) (x+3) (x+g) (x+g+1) (x+g+2) (x+2g) (x+2g+1) (x+3g)
If I add these together I will get 10x+10g+10.
This is a random 4 step stair from a ten by ten grid. I could also write this algebraically like:
(x) (x+1) (x+2) (x+3) (x+g) (x+g+1) (x+g+2) (x+2g) (x+2g+1) (x+3g)
If I add these together I will also get 10x+10g+10.
These results show that the equation works for 8 by 8 and 10 by 10 grids. I will now test it on a 6 by 6 grid to prove if it works or not.
I am going to take a random 4 step stair and test the equation.
5 step stairs on different size grids
I will now take a random 5 step stair from the 8 by 8 grid.
I will also take a random 5 step stair from a 10 by 10 grid.
15x+20g+20= stair total for 5 step stairs.
I will now test it on a random 5 step staircase from a 6 by 6 grid.
6 step stairs.
I will start by taking a random 6 by 6 step stair. I can see that x = 10. If I write this out algebraically I would get: (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+3g) (x+3g+1) (x+3g+2) (x+4g) (x+4g+1) (x+5g)
I will now take a random 6 step stair from a 10 by 10 grid.
If I wrote this out algebraically I find that it would be the same as in the 8 by 8 grid being: (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+3g) (x+3g+1) (x+3g+2) (x+4g) (x+4g+1) (x+5g).
If I simplify all this I would find that the equation would be 21x + 35g + 35. This represents the equation for a 6 step stair on any size grid.
I will now test the equation on a random 6 step stair from a 6 by 6 grid.
x=1 and g=6. If I substitute these into the formula I will get (21 multiplied by 1) + (35 multiplied by 6) add 35. This will equal 266. If you add the numbers individually you will also get 266 which shows that the equation works.
Here is a table of my results
I will now try and find an equation for any size stair on any size grid. The final product will be an equation like this: (part1)x + (part2)g + (part3).
