In my results I noticed that.
- All the numbers are even numbers;
- When the staircase is moving up the number 60 is added each time to the next number;
- When the staircase is moving diagonally the number 66 is added each time to the next number;
- The two sequences don’t start with one.
So I discovered that two sequences were created. But, to use these sequences I had to find a formula which by knowing the bottom left number (this number has to be in the first column when the staircase is moving up, or diagonally) of the staircase I could work out the total. To do this, I used an nth term formula.
Staircase moving upwards: Staircase moving diagonally:
-10 50, 110, 170, … -16 50, 116, 182, …
So the formula is: So the formula is:
60n - 10 66n - 16
By testing the two formulae with numbers only in the first column and going diagonally, I discovered that the two don’t work and don’t give you the total sum of the numbers in the staircase when you substitute n with a number. Instead, I found out that if you insert the total sum of the numbers in a staircase (with its bottom left number anywhere on a 10x10 grid), in the formula for the staircase moving upwards, n is equal to the row number in which the bottom left number of the chosen staircase is in; whilst for the formula of the staircase moving diagonally, if you insert in it the total sum of numbers in a staircase which its bottom left number is in the diagonal line, n is equal to the position of the bottom left number in the diagonal line (e.g. 1st, 2nd, 3rd, etc… see table in next page, the number in blue is the number of which I worked out the position).
Once I found these two formulae I made a set of predicted results.
Example of staircase moving up:
Bottom left number of staircase = 45 Row number: 60n – 10 = 341
Total sum = 6n + 44 60n = 354
= (6 x 45) + 44 n = 354 ÷ 60
= 270 + 44 n = 5.9
= 341
In fact the number 45 is in the 5th row:
Example of staircase moving diagonally:
Bottom left number = 34
Total sum = 6n + 44 Position of the number: 66n – 16 = 248
= (6 x 34) + 44 66n = 264
= 204 + 44 n = 264 ÷ 66
= 248 n = 4
In fact the number 34 is the fourth in the diagonal line:
So this shows that the two formulae I found could be used to either find the row in which a number is in, or, if that number is in the diagonal line you can find out exactly at which point.
Lastly, I wanted to try to investigate more the relation between the total sum of the numbers in a stair case, and its bottom left number: if I double the bottom left number will also the total sum double? Are the two directly proportional?
Bottom left number = 4 Bottom left number = 8
Total sum = 6n + 44 Total sum = 6n + 44
= (6 x 4) + 44 = (6 x 8) + 44
= 24 + 44 = 48 + 44
= 68 = 92
Bottom left number = 22 Bottom left number = 44
Total sum = 6n + 44 Total sum = 6n + 44
= (6 x 22) + 44 = (6 x 44) + 44
= 132 + 44 = 264 + 44
= 176 = 308
Concluding, the bottom left number of a staircase is not directly proportional to the total sum of the number in it. The two factors are proportional because as one increases also the other does, but they are not directly proportional because as one doubles, the other doesn’t double exactly.
AIM: INVESTIGATE FURTHER RELATIONSHIP BETWEEN THE STAIR TOTALS AND OTHER STEP STAIRS ON OTHER NUMBER GRIDS
Given any times table, and a staircase of any steps, I tried to investigate further relations between the two, to conclude with a general formula.
I thought it was a good idea to start with a table with on one side the staircases with various numbers of steps, and on the other side the various sizes of the grid. I thought that I could start to do this firstly to see if any sequence would come out, and secondly to start to get some more general formulae (e.g. a formula for a 3 step staircase but for any grid).
Here is part of the table:
No. of steps of staircase
Size of
Grid
*The formulae highlighted in green theoretically would not exist as the staircase is bigger than the table, but I’ve included them as they are part of the pattern
To fill in this table I simply used the various times table but in algebraic form (like the ones in part 1). Then I highlighted the staircase with the chosen numbers of stairs, and I added up what was in each highlighted part. Here are two examples:
4 steps, on a 8x8 grid 3 steps, on a 4x4 grid
x+x+1+x+2+x+3+x+8+x+9+x+10 x+x+1+x+2+x+4+x+5+x+8
+x+16+x+17+x+24 = 10n + 90 = 6n + 20
After, I also calculated some more general formulae by keeping the first part of the formula the same, and then adding the difference of the numbers above and the difference between the size of the grid and the numbers above:
So, once I completed the table I was immediately looking for some results, and I noticed that many sequences came out.
Firstly, I noticed a sequence which was in every line and it appeared with n: 3 + 6 + 10 + 15 + 21… After a bit of research I found out that this is a quadratic sequence, made of triangular numbers; and I also discovered through some research in books that the general formula for quadratic equations is an2 + bn + c, where a, b, and c are the differences in the formula:
2nd term 3rd term 4th term etc….
1st term
1 3 6 10 15 21 Sequence
+2 +3 +4 +5 +6 1st difference
+1 +1 +1 +1 2nd difference
Now, through proving that this formula works (by substituting “n” with the position of the term in the sequence e.g. 1st, 2nd, 3rd, etc… and seeing if the right term comes out), using substitution and using simultaneous equations, I am going to discover what a, b, and c are so that I can come up with a formula for this sequence.
Substituting “n” with the term of the formula, to find “c”
n = 1 n = 2 n = 3
an2 + bn + c an2 + bn + c an2 + bn + c
a x 12 + b x 1 + c a x 22 + b x 2 + c a x 32 + b x 3 + c
a + b + c = 1 4a + 2b + c = 3 9a + 3b + c = 6
Now that I am sure that the formula works I can work out what “c” is from rearranging and making “c” the subject of the formula when I substituted n with 1. I am finding out “c” first because it is the only constant difference, so before working out the other differences (a and b) I have to work out c.
c = 1 – a – b
Substituting “c” into the equations to find “b”
Substituting c into n=2 equation Substituting c into n=3 equation
4a + 2b + c = 3 9a + 3b + c = 6
c = 1 – a – b c = 1 – a - b
4a + 2b + 1(1 – a – b) = 3 9a + 3b + 1(1 – a – b) = 6
4a + 2b + 1 – a – b = 3 9a + 3b + 1 – a – b = 6
3a + b = 2 8a + 2b = 5
b = 2 – 3a
Using simultaneous equations with the two formula for “b” to find a
b = 2 – 3a
8a + 2b = 5
8a + 2(2 – 3a) = 5
8a + 4 – 6a = 5
a = 1
2
So, because a = 0.5, then:
b = 2 – 3a c = 1 – a - b
= 2 – 1½ c = 0
b = 1
2
This finally gives us the formula for this equation, which is: ½n2 + ½n
But, because this is a formula which finds the nth term, so you have to input into the formula the number of term, then the formula becomes:
Where: “n” = bottom left number of staircase, e.g 3
n(½s2 + ½s) “s” = number of stairs in staircase e.g
3 stairs = 3
So, looking at my table in more depth I found two cubic sequences:
0, 11, 44, 110, 220, … Sequence
+11 +33 +66 +110 1st difference
+22 +33 +44 2nd difference
+11 +11 3rd difference
1st term 2nd term 3rd term 4th term …
0, 1, 4, 10, 20, 35, … Sequence
+1 +3 +6 +10 +15 1st difference
+2 +3 +4 +5 2nd difference
+1 +1 +1 3rd difference
To find my general formula which would tell me the total sum of the numbers in any staircase and in any size grid I decided to use the second cubic sequence. This is firstly because after some research I discovered that they both make you end up with more or less the same formula; and secondly because I noticed that the terms in the second sequence are equal to the ones in the first sequence ÷ 11. So if I would use the other one I would have make one calculation more.
Using some books and by asking my teacher I found out that the general formula for cubic sequences is an3 + bn 2 + cn + d. And now to find the nth term formula for this sequence I will do the same thing as I did for the quadratic sequence: I will use substitution and other methods to work out a, b, c and d so that I will be able to find a formula.
Substituting n with the first four terms of the sequence to find “d”
n = 1 a x 13 + b x 12 + c x 1 + d
a + b + c + d = 0
n = 2 a x 23 + b x 22 + c x 2 + d
8a + 4b + 2c + d = 1
n = 3 a x 33 + b x 32 + c x 3 + d
27 a + 9b + 3c + d = 4
n = 4 a x 43 + b x 42 + c x 4 + d
64a + 16b + 4c + d = 10
These proves that the general formula for cubic sequences is right.
To find “d” I will rearrange the formula n = 1.
d = -a –b –c
Substituting “d” into n=2 and n=3 formula to find “c”
8a + 4b + 2c + d = 1
8a + 4b + 2c + (-a –b –c) = 1
8a + 4b + 2c – a – b – c = 1
7a + 3b + c = 1
c = 1 – 7a – 3b
27a + 9b + 3c + d = 4
27a + 9b + 3c (-a –b –c) = 4
26a + 8b + 2c = 4
Substituting “c” into n = 3 equation to find “b”
Now I am going to substitute “c” in the n=3 equation which I used to substitute “d”, and not in the previous one, because the last one I used can be considered as more completed because “d” has already been calculated from it.
26a + 8b + 2c = 4
26a + 8b + 2(-7a -3b +1) = 4
26a + 8b – 14a – 6b + 2 = 4
12a + 2b + 2 = 4
12a + 2b = 2
2b = 2 – 12a
b = 1 – 6a
Finding “a”
To find the value of “a” I found an alternative method other than substitution, which gives me the right answer.
This is the “normal” cubic sequence made of cubic numbers:
1, 8, 27, 64, 125 Sequence
+7 +19 +27 +61 1st difference
+12 +13 +14 2nd difference
+6 +6 3rd difference
As you may have noticed, the terms in the third difference of the “normal” cubic sequence are 1/6 of the terms in the third difference of the cubic sequence I used. Therefore this will bring us to conclude that:
a = 1
6
Now, even if I have the value of all four letters they are not in their simplest form. So now I’ll simplify their values also because it will then be easier for me to solve long equations as the value of letters will be shorter.
d = - a – b – c
c = 1 – 7a – 3b
b = 1 – 6a
a = 1
6
(this also proves that the value that I found for “a” is corrcect)
b = 1 – 6/1 x 1/6
b = 1 – 1
b = 0
c = 1 – 7a – 3b
c = 1 – 7/1 x 1/6 – 3 x 0
c = 1/1 – 7/6 – 0
c = 6 - 7
6
c = -1/6
d = -a –b –c
d = -1/6 -1/6 - 0
d = 0
So now that I have the values of the four letters I can substitute the letters in the general cubic sequence with the specific values, and then I will discover the formula to find any term of this sequence.
an3 + bn2 + cn + d
1/6 n3 + 0n2 – 1/6 n + 0
= 1/6 n3 – 1/6 n
This is the formula of the cubic sequence. So, if for example now I want to know what the fourth term of the sequence is, I would have to substitute n with 4. Now I will test this formula:
n = 2
= 1/6 x n3 – 1/6 x n
= 1/6 x 23 – 1/6 x 2
= 4/3 – 1/3
= 3/3
= 1
n = 3
= 1/6 x n3 – 1/6 x n
= 1/6 x 33 – 1/6 x 3
= 1/6 x 27/1 – 1/6 x 3/1
= 9/2 – 3/2
= 6/2
= 3
The results show that the formula for the cubic sequence is right as they are equal to the actual terms in the sequence.
Now I finally have all the things needed to make a formula which finds the total of any step staircase in any size grid. This formula will be made by:
(Formula for quadratic sequence) + formula for cubic sequence +
g(formula for cubic sequence)
The second time I am adding the formula for the cubic sequence x “g” because if you look back at the general formulae in the table, there’s a number x g + the same number without g; so now in my final formula I just have to repeat the same thing.
So, the three formulae are:
n(½s2 + ½s) and 1/6n3 – 1/6 n and g(1/6n3 – 1/6 n)
To make it work I have to substitute “n” with “s” because I will insert the bottom left number (“n”) only at the beginning of the formula as I only need it once, and also because now I have to discover the total sum of numbers in a staircase, not any more a term of the cubic sequence.
This gives me the final formula:
n(½ s2 + ½ s) + g (1/6 s3 – 1/6 s) + (1/6 s3 – 1/6 s)
where:
n = bottom left number
s = number of steps in staircase
g = size of grid e.g. 10x10 = 10
As a last thing I will test the formula to show that it works.
n = 1
s = 3
g = 10
Total sum should be = 50
Total sum = n(½ s2 + ½ s) + g (1/6 s3 – 1/6 s) + (1/6 s3 – 1/6 s)
= 1(1/2 x 9 + ½ x 3) + 10 (1/6 x 27 – 1/6 x 3) + (1/6 x 27 -1/6 x 3)
= 1(9/2 + 3/2) + 10 (9/2 – ½) + (9/2 – ½)
= 12/2 + 10(8/2) + 8/2
= 6 + 10 x 4 + 4
= 6 + 40 + 4
= 50
This proves and shows that my formula is correct and that it can calculate the total sum of the numbers inside any step stair in any size of grid.
