An example of the formula T=T(t-1)+6 working in practice is the calculation for N=54,
T(t-1)=The total for the number N=53 i.e. 362.
T=362+6
T=368 where N=54
I checked my answer was correct by adding up the step stair N=54, 54+55+56+64+65+74=368.
This shows that the formula works for N=54.
To check the formula I am also going to use N=78 as an example,
T(t-1)=The total for the number N=77 i.e. 506
T=506+6
T=512 where N=78
I checked my answer was correct by adding up the step stair N=78,
78+79+80+88+89+98=512
This shows the formula works for N=78
Now for confirmation I will work out the total of N=15,
T(t-1)=The total for the number N=14 i.e. 128
T=128+6
T=135 where N=15
I checked my answer was correct by adding up the step stair N=15,
15+16+17+25+26+35=135
This shows the formula works for N=15
This shows that each total on the horizontal row is 6 more than the previous total, except the totals in the 1, 9 and 10 vertical row and 9 and 10 horizontal row. The formula to show this is T=T(t-1)+6.
I am now proceeding to investigate vertical columns; I am looking at the totals ad comparing them as they proceed up the row.
1=50
60
11=110
60
21=170
60
31=230
60
41=290
60
51=350
60
61=410
60
71=470
I could do not do 81 or 91 because using either as the numbers as N I could not complete the 3 step stair.
I am now going to investigate the second column,
2=56, 12=116, 22=176, 32=236, 42=296, 52=356, 62=416, 72=476
I could immediately see from this that the difference between the totals is 60.
I am now going to investigate the third column,
3=62, 13=122, 23=182, 33=242, 43=302, 53=362, 63=422, 73=482
I am now going to investigate the fourth column,
4=68, 14=128, 24=188, 34=248, 44=308, 54=368, 64=428, 74=488
I am now investigating the fifth column,
5=74, 15=134, 25=194, 35=254, 45=314, 55=374, 65=434, 75=494
I am now going to investigate the sixth column,
6=80, 26=140, 36=200, 46=260, 56=320, 66=380, 76=440
I am now investigating the seventh column,
7=86, 17=146, 27=206, 37=266, 47=326, 57=386, 67=446, 77=506
I am now investigating the eighth column,
8=92, 18=152, 28=212, 38=272, 48=332, 58=392, 68=452, 78=512
From this I can see that a formula that could arrive at the correct total is
T=T(t-1)+60
I arrived at this formula when I saw that the total can be produced by taking the total before it and adding 60 to it. This formula works on all the numbers except those specified above and the bottom horizontal row because they do not have a preceding total. You can also see that the step stair consists of 6 numbers each of these numbers moves 1 position above themselves becoming greater by 10. When all 6 of the numbers become greater by 10 the total increases by 60, 10x6=60 the total becomes greater by 60.
I am now going to give an example of the formula working,
I will workout N=38,
T=212+60 T=272. This is correct and shows that the formula works, if you take the total of the number preceding N and plus 60 to it, you arrive at the total of the N number you are trying to work out.
I will check my formula by testing it again by working out N=14,
T=68+60 T=128. Once again my formula has worked out the correct answer. This shows my formula still works for N=14.
I will now finally check my formula by find the answer to N=23,
T=122+60 T=182. This shows my formula works because I have checked it three times.
This shows that each total on the vertical column is 60 more than the previous total, except the totals in the 9 and 10 vertical columns and 9 and 10 horizontal row. The formula to show this is T=T(t-1)+60.
I have now decided to investigate the total for the rows and columns by adding up the totals in the rows.
Horizontal rows
I am investigating the first row,
50 + 56 + 62 + 68 + 74 + 80 + 86 + 92 = 568
I am now investigating the second row,
110+116+122+128+134+140+146+152=1048
I am now investigating the third row,
170+176+182+188+194+200+206+212=1528
From the first 3 rows I can see that the totals of the rows go up by 480 each time.
I will now investigate the fourth row,
230+236+242+248+254+260+266+272=2008
I will now investigate the fifth row,
290+296+302+308+314+320+326+332=2488
I will now investigate the sixth row,
350+356+362+368+374+380+386+392=2968
I will now investigate the seventh row,
410+416+422+428+434+440+446+452=3448
I will now investigate the seventh row,
470+476+482+488+494+500+506+512=3928
From the sums above I can clearly see the total of the previous row is 480 more than the preceding row. From this I can see that a formula that would give the correct answer would be:
R=R(r-1)+480
I produced this formula when I saw that to arrive at a total of a row you have to take the total of the preceding row and add 480 to it. This works for all rows except the first row.
I am now testing to see whether my formula works,
I will calculate the total of row 5,
R=2008+480 R=2448
My formula has produced the correct total, this shows it works, I will check it again.
I will the total of row 2,
R=568+480 R=1048,
Once again my formula has worked.
I will now work out the total for row 4,
R=1528+480 R=2008.
This proves my formula for producing row totals works.
This shows that to arrive at the total for a horizontal row you add 480 to the previous row total, except the totals in the first horizontal row. The formula to show this is R=R(r-1)+480.
Vertical Column Totals
I will investigate the first column,
50+110+170+230+290+350+410+470=2080
I will investigate the second column,
56+116+176+236+296+356+416+476=2128
I will now investigate the third row,
62+122+182+242+302+362+422+482=2176
From this I can see that the difference between the totals of the columns is 48, I will carry on my investigation to check.
I will now investigate the fourth column,
68+128+188+248+308+368+428+488=2224
I will now investigate the fifth column,
74+134+194+254+314+374+434+494=2272
I will now investigate the sixth column,
80+140+200+260+320+380+440+500=2320
I will now investigate the seventh column,
86+146+206+266+326+386+446+506=2368
I will now investigate the eighth column,
92+152+212+272+332+392+452+512=2416
From the sums above I can clearly see the total of the previous column is 48 more than the preceding column. From this I can see that a formula that would give the correct answer would be:
C=C(c-1)+48
I produced this formula when I saw that to arrive at a total of a column you have to take the total of the preceding column and add 48 to it. This works for all columns except the first column and the ninth and tenth columns.
I am now testing to see whether my formula works,
I will calculate the total of column 6,
C=2272+48 C=2320.
My formula has produced the correct total, this shows it works, I will check it again.
I will the total of row 8,
C=2368+48 C=2416,
Once again my formula has worked.
I will now work out the total for row 4,
C=2176+48 C=2224.
This proves my formula for producing row totals works.
This shows that to arrive at the total for a vertical column you add 48 to the previous column total, except the totals in the first column and the ninth and tenth columns. The formula to show this is C=C(c-1)+48.
I made a table to show my results in working out the totals for the step stairs and the rows and columns.
The Step Stair
I am now going to investigate ways of arriving at the total by using numbers in the 3 step stair.
Within the table above are the letters I will use to refer to the points in the 3 step stair.
I am going to see if I can use N to arrive at the total for the step stair,
I will start by using them in the number stair N=25,
My hypothesis is that the formula to establish the total of a 3 step stair is T= 6N+44. To arrive at this I have already tried two alternatives (see Appendix 1). I will now test using the above formula. T= (6x25)+44. Therefore T=194 which agrees with the above table.
I will now check my formula to see if it works with N=13,
T= (6x13)+44, T=122,
I will check if my answer is correct, 13+14+15+23+24+33=122.
The formula works for N=13
Further checks:
N=52,
T= (6x52)+44=356,
The answer is correct as 52+53+54+62+63+72=356.
N=33,
T= (6x33)+44=242
33+34+35+43+44+53=242.
The formula works with N=33 as well.
That means the formula T=6N+44 is correct and will give the answers for any 3 step stair on a 10 by 10 grid.
The formula links both the position on the grid, N and the total of the step stair, T in one formula. Another way to phrase the formula would be, T=N+(N+1)+(N+2)+(N+10)+(N+11)+(N+20). I will demonstrate this in a table below:
These are three, 3 step stairs they demonstrate how I got the formula above, table A represents a step stair numerically, table b represents how to get the other numbers in the step stair from using the N and table c shows what will happen when the formula combines. Table b shows 6 N’s each representing the N number in the step stair whether it is 25 or 78, the +10 or +2 part of the formula represents what has to happen to the N number so that it can become that position in the step stair. If you count up the N’s there are 6, 6N and if you add up all the extra numbers in the table they come to 44, this gives the formula 6N+44, from its non-cancelled down form.
Part 2
Introduction
I am going to investigate further relationships between stair totals and other step stairs on other number grids.
Expressions
I am going to use these letters to show certain numbers:
N= bottom left hand number on the step stair, the position
T= total of the step stair
T(t-1)= total of step stair before
G= the grid size
S=the number of squares in a step stair grid
A= the total, T, of the step stair N=1-SN
3 Step Stairs on Other Sized Grids
I am going to investigate 3 step stairs on a 6 by 6 grid, below:
Data Collection
The grid below summarises the totals of the step stairs, and is identical to the grid above except the N numbers are replaced by the totals for that position,
I can see that the difference between the N numbers as they proceed along the row is still 6, this means the formula T=T(t-1)+6 still applies to the 6 by 6 grid. This is because each step stair consists of 6 numbers, each of these numbers moves 1 position to the right, this means the step stair before becomes greater by 6 because each of the 6 positions in the step stair move 1 place to the right, 1x6=6 which means the step stair becomes greater by 6.
But as they proceed up the column the difference has changed from 60, to 36, that means the formula to work out the total of the step stair above a step stair is T=T(t-1)+36, because the difference between the step stairs has changed so does the latter number in the formula.
I will now try a variation of the formula 6N+44 to try to work out a relationship between the position on the 6 by 6 grid and the total of the step stair, the formula is,
SN+A.
In the above formula the S=the number of squares in a step stair grid, the N= the bottom left hand step stair position and the A= the total, T, of the step stair N=1-SN.
This formula is universal, it will work on any square grid size with any sized step stair and I will prove it.
Firstly I will check if the formula works on this grid,
I will fill in the letters with numbers,
SN+A becomes
6N+28,
The 6 is the number of steps in the step stair, the N will be replaced by the N number of the step stair, and the 28 is the total of the step stair N=1 i.e. 34 (1+2+3+7+8+13=34) minus S multiplied by N i.e. 6x1=6, so A=34-6, A=28.
The formula for the any total on the 6 by 6 grid is now 6N+28.
I will check to see if my formula is correct,
I will use my formula to find the total of N=7,
The formula will be,
6x7+28=70
I will check if my answer is correct 7+8+9+13+14+19=70,
This shows that the formula 6N+28 works for N=7.
I will check my formula again by working out the answer to N=21,
The formula will be,
6x21+28=154
I will check the answer is correct, 21+22+23+28+29+34=154
This shows that the formula works for N=21.
I will now do a final check to see if my formula works by working out the answer to N=15,
The formula will be,
6x15+28=118,
I will now check to see if the answer is correct, 15+16+17+21+22+27=118,
This shows the formula works for N=15,
And the formula 6N+28 works on a 6 by 6 grid.
I am now going to investigate 3 step stairs on a 9 by 9 grid,
Data Collection
The grid below summarises the totals for the step stairs above,
I can see that the difference between the N numbers as they proceed along the row is still 6, this means the formula T=T(t-1)+6 still applies to the 6 by 6 grid. This can be changed to apply to any step stair on any grid, the new formula is,
T=T(t-1)+S.
This is because each step stair consists of 6 numbers, each of these numbers moves 1 position to the right, this means the step stair before becomes greater by 6 because each of the 6 positions in the step stair move 1 place to the right, 1x6=6 which means the step stair becomes greater by 6. And if the size of the step stair changes the S changes.
But as they proceed up the column the difference has changed again from 36, to 54. From this I can see a recurring pattern that applies to 3 step stairs. The numbers of positions in a step stair multiplied by the grid size, SG, will give the difference between the number above and the number below,
e.g. To find the number f