52-12=
40=
Testing my formula
Looking at my table when n= 4, T= 64
I can now use my formula which I have just found to check if it is correct
T= 6(4) + 40
T= 24 + 40
T= 64
This shows that my formula works
Proving my formula
I have noticed that on my number stairs, as you go across the number increases by 1 and as you go up the number increases by 9. I will illustrate this below.
+ 9
+ 9
+ 1 + 1
I will now do the same thing using algebra.
+ 1
+ 9
+ 9
+ 1 + 1
T= n+ (n+ 1) + (n+ 2) + (n+ 9) + (n+ 10) + (n+ 18)
T= 6n+ 40
I will now look at a 3 step stair on an 8 by 8 Grid
Analysis
n= 1
T= 1+ 2+ 3+ 9+ 10+ 17
T= 42
n= 2
T= 2+ 3+ 4+ 10+ 11+ 18
T= 48
n= 3
T= 3+ 4+ 5+ 11+ 12+ 19
T= 54
n= 4
T= 5+ 6+ 7+ 13+ 14
T= 60
n= 5
T= 5+ 6+ 7+ 13+ 14+ 21
T= 66
I will now draw a table because it is at a glance data, which means that it is quick and easy to look at, and furthermore drawing a table will help me find the nth term later on.
+ 6 + 6 + 6 + 6
Observations
From the table I have noticed a few trends which are as follows;
- All the stair Totals(T) are even numbers
- The stair Totals(T) increase by 6 each time
Finding the nth term
T= 6n +
From the table I can say that When n= 5, T must be 66
T= 6 x 5+
66= 30+ This means that
66- 30=
36=
Testing the formula
Looking at my table when n= 2, T must = 48
I can now use my formula which I have just found to check if it is correct
T= 6(2) + 36
T= 12 + 36
T=48
This shows that my formula works
Proving my formula
I have noticed that on my number stairs, as you go across the number increases by 1 and as you go up the number increases by 8. I will now illustrate this below.
+8
+8
+1 +1
I'll now do the same thing using algebra.
+ 1
T= n+ (n+ 1) + (n+ 2) + (n+ 8) + (n+ 9) + (n+ 16)
T= 6n+ 36
Relationship between stair totals
I will now find the link between the stair totals only for 3 step stairs. I will also look for the relationship between the stair totals and the grid size.
I will now draw a table because it is at a glance data, which means that it is quick and easy to look at because the data is grouped together, and furthermore it will help me to look for patterns in the data.
Prediction
6n+ (4G) + 4
I now have a main formula for a 3 step stair on a G by G grid (any grid size).
Testing the formula
For a 11 by 11 grid If G= 11 and n= 4,
T= 6(4) + (4x 11) + 4
T= 24 + 44 + 4
T= 72
I will now draw a diagram of my 3 step stair to ensure that the stair totals are the same
n= 4
T= 4 + 5 + 6 + 15 + 16 + 26
T= 72
Proving my formula
I have noticed that for any 3 step stairs, as you go across you add 1, and as you go up it increases by the grid size (G).
+1
+ G
+G
+1 +1
T= n+ (n+ 1) + (n+ 2) + (n+ G) + (n+ G+ 1) + (n+ 2G)
T= 6n+ 4G+ 4
Prediction
Now that I have proved that my formula is correct, I can now use this formula to make predictions.
For a 14 x 14 grid
If G= 14 and n= 7
T= 6(7) + 4(14) + 4
T= 42 + 56 + 4
T= 102
I will now draw a 3 step stair to make sure the stair totals are the same
n=7
T= 7 + 8 + 9 + 21 + 22 + 35
T= 102
Examining other stair sizes
I have now found the formula for any 3 step stair, so I will now investigate other stair sizes.
To be systematic I will start by looking at stair size 1 and work my way upwards to bigger stair sizes.
Stair Size 1
If the grid size (G) = 12 by 12 and n= 6, then T= 6
In this formula there is no grid size involved.
I will now look at a 2 step stair
Because I have seen the pattern that as you go across you add 1 and as you go up you add the grid size, I will now use this concept throughout my coursework.
I will now look at a 2 step stair
T= n + (n+ 1) + (n+ G)
+ G
+ 1
Testing my 2 step stair formula
If G= 8 and n=3
T= 3(3) + 8 + 1
T= 9 + 8 + 1
T=18
I'll now draw the 2 steps stair to ensure that the stair totals are the same
T= 3 + 4 + 11
T= 18
I will now look at a 4 step stair
I'll now look at the 4 step stair.
+ G
+ G
+ G
+1 +1 +1
T= n+ (n+1) + (n+2) + (n+3) + (n+G) +(n+G+1) + (n+G+2) + (n+2G) + (n+2G+1) + (n+3G)
Testing the 4 step formula
If G= 9 and n= 5
T= 10(9) + 10(5) + 10
T= 90 + 50 + 10
T=150
I will now draw the 4 step stair to make sure the stair totals are the same
+ 9
T= 5 + 6 + 7 + 8 + 14 + 15 + 16 + 23 + 24 + 32
T= 150
+ 9
+ 9
+ 1 + 1 + 1
I will now look at a 5 Step Stair
+G
+G
+ G
+G
+1 +1 +1 +1
T= n + (n+1) + (n+2) + (n+3) + (n+4) + (n+G) + (n+G+1) + (n+G+2) + (n+G+3) + (n+2G) +
(n+2G+ 1) + (n+2G+2) + (n+3G) + (n+3G+1) + (n+4G)
Testing the 5 step stair formula
If G= 12 and n= 6
T= 15(6) + 20(12) + 20
T= 90 + 240 + 20
T= 350
I will now draw the 5 step stair to ensure that the stair totals are the same
+ 12
+ 12
+ 12
+ 12
+ 1 + 1 + 1 + 1
T= 6+ 7+ 8+ 9+ 10+ 18+ 19+ 20+ 21+ 30+ 31+ 32+ 42+ 43+ 54
T= 350
I will now put my results into a table where I will compare the stair size with the formula for the stair total (T)
I will use x to represent the stair size
Prediction
Observations
- I have noticed that the number next to the co-efficient (n) increases by 1 each time
- The coefficient of n is a triangular number, I will find out the formula for the triangular numbers later on
-
The 2nd and the 3rd terms in the formulas are always the same
-
For the 2nd and the 3rd terms, triangular numbers are added to find the next term
I will now try to find a general formula involving the 3 terms.
T = n + G +
So if I had stair size 5
T = (5th triangular number)n + (sum of 1st 4 triangular number)G + (Sum of 1st 4 triangular
numbers)
T = 15n + (1 + 3 + 6 + 10)G + (1 + 3 + 6 + 10)
T = 15n + 20G + 20
This is the same formula as shown in the table on page 18
I now have a formula that is in words, but I would much prefer to have a formula using algebra.
Finding a formula for triangular numbers
Difference 2 3 4
1 1
I will now substitute numbers into the formula
When x =1 a(1)2 + b(1) + c = 1
a + b + c = 1
When x = 2 a(2) + b(2) + c = 3
4a + 2b + c = 3
When x = 3 a(3) + b(3) + c = 6
9a + 3b + c = 6
Solving Simultaneously
3 9a + 3b +c = 6
Subtract
2 4a + 2b + c = 3
5a + 1b = 3
2 4a + 2b + c = 3
Subtract
1 a + b + c = 1
5 3a + 1b = 2
Using 4 + 5
5a +1b = 3
Subtract
3a +1b = 2
2a = 1
a = 1
2
Use 4
5a + 1b = 3
5 ( 1 ) + b = 3
2
2 1 + b = 3
2
b = 1
2
Use
a + b + c = 1
1 + 1 + c = 1
2 2
1 + c = 1
c = 0
So therefore ax2 + bx + c
is 1 x2 + 1 x + 0
2 2
=1 x2 + 1 x I will factorize this
2 2
So the formula for a triangular number is
T= (Triangular number) n + (sum of triangular numbers but 1 below stair size)G + (Sum of triangular numbers but 1 below stair size)
T= ( 1 x(x + 1))n +(Σ triangular numbers less 1)G + (Σ triangular numbers less 1)
2
I will now check my formula
Stair size of 4
x = 4
T= ( 1 (4) (4+1)] n + [1+3+6] G + [1+3+6]
2
T= 10n + 10G + 10
Linking the stair size with triangular numbers
Triangular
Numbers
1 3 6 10
Stair
Shape
I can now rearrange my stairs to make them look like triangles
1 1+2 1+2+3 1+2+3+4
= 1 = 3 = 6 = 10
Proving the Triangular number Formula
I have discovered that the triangular numbers can be found by
1 x (x+1)
2
But I want to show why
To show this I will now look at the 4th triangular number.
1 3 6 10
1 + 2 + 3 + 4
4 + 3 + 2 + 1 Rewrote backwards
5 + 5 + 5 + 5
So 4 x 5 = 20 I need to find 10
So in order to find 10, I’ll do
20 = 10
2
I will now use the same principle find it out using algebra.
1 + 2 + 3 +………………x-1 + x
+
x + x – 1 + x – 2 +……… 2 + 1
(x+1) + (x+1) + (x+1)…. (x+1) + (x+1)
x lots of x+1
=x(x+1)
I will now divide by 2, so
x (x+1) This is the same formula that I found out
2 earlier, this proves that my formula works
Refining my General formula This is my formula I found out earlier
T= ( x ( x + 1) ) n + (Σ triangular numbers less 1) G + (Σ triangular numbers
2 less 1)
x-1 x-1
T= ( x ( x + 1) ) n + (Σ x (x+1) ) G + (Σ x (x+1) )
2 x=1 2 x=1
This sorts out the ‘ less 1’ stated previously
I have added my formula for
Triangular numbers
Checking the formula
x=3 n=2 G=9
T= (3( 3+1 ) ) + (1(1+1)+ 2(2+1) ) 9 + (1(1+1)+ 2(2+1) )
2 2 2 2 2
T= 6(2) + ( 1 + 3 )9 + ( 1 + 3 )
T= 12 + 4(9) + 4
T=12 + 36 + 4
T= 52
I will now draw the 3 step stair to check if the stair totals (T) are the same.
+ 9
+ 9
+ 1 + 1
T= 2 + 3 + 4 + 11 + 12 + 20
T= 52
Improving the formula
I want to see if I can find a better way to make the 2nd and 3rd terms.
The 2nd term has coefficients.
x=1 x=2 x=3 x=4 x=5
0 1 4 10 20
1 3 6 10
2 3 4
1 1
1st difference
2nd difference
3rd difference
If it was the second difference, than I would be a quadratic formula but because it is the 3rd difference, I would expect a cubic formula.
Cubic formula= ax3 + bx2 + cx + d
I will now try to find the a, b, c and the d value by substituting x=1, x=2, x=3, x=4
x = 1 x = 2 x = 3 x = 4
0= a+b+c+d 1= 8a+4b+2c+d 4=27a+9b+3c+d 10=64a+16b+4c+d
Difference
1=7a+3b-c 3=19a-15b+c 6=37a+7b+c
Difference
2= 12a + 2b 3=18a+2b
Difference
1 = 6a
1 = a
6
I am now going to find the b value by substituting
a = 1 into 2= 12a +2b
6
2= 12 (1/6) + 2b
2= 2 + 2b
0= 2b
0= b
I am now going to find out my c value by substituting
a and b into 1= 7a + 3b +c
1= 7(1/6) + 3(0) + c
1= 7 + c
6
-1 = c
6
I will now find d by substituting a, b and c.
0= 1 + 0 + -1 + d
6 6
0= d
I'll now use my a , b ,c and d values to find my cubic formula.
Cubic formula= ax3 + bx2 + cx + d
=1x3 + 0x2 - 1 + 0
6 6
=1x3 - 1x I will now factorize this
- 6
=1x(x2-1)
6
I will now check my formula by using x = 3
When x=3
1(3) (32-1)
6
= 3 (8)
6
= 1 (8)
2
= 4
I will now replace the 2nd and the 3rd term in my old formula with the new formula which I have found
The old formula is
T= (x(x+1)) n + (Σ x(x-1)) G + (Σ x(x-1))
2 2 2
The new one is
T= (x(x+1)) n + (Σ1(x2-1)) G + 1x(x2-1)
2 6 6
I will now check it using the stair size 5.
x= 5(stair size) n=2(stair number) G= 8(Grid size)
T= 5(6)(2) + (1 (5)(52-1))8 + 1(5)(52-1)
2 6 6
T= 30 + 5(24)(8) + 5(24)
6 6
T= 30 + 20(8) + 20
T= 30 + 160 +20
T= 210
I will now do a diagram of the stair, to make sure the stair totals ( T ) are the same
T= 2 + 3 + 4 + 5 + 6 + 10 + 11 + 12 + 13 + 18 + 19 + 20 + 26 + 27 + 34
T= 210
Now that I have completed my coursework, I will outline the formulas that I have found during this coursework, and what it does.
Formula _ What it does
T= 6n + 44 Formula for any 3 step stair
on a 10 by 10 grid
T= 6n + 40 Formula for any 3 step stair
on a 9 by 9 grid
T= 6n + 36 Formula for any 3 step stair
On a 8 by 8 grid
T= 6n + 32 Formula for any 3 step stair
On a 7 by 7 grid
T= n Formula for any 1 step stair
On any grid size
T= 3n + G + 1 Formula for any 2 step stair
On any grid size
T= 6n + 4G + 4 Formula for any 3 step stair
On any grid size
T= 10n + 10G + 10 Formula for any 4 step stair
On any grid size
T= 15n + 20G + 20 Formula for any 5 step stair
On any grid size
T= 21n + 35G + 35 Formula for any 6 step stair
On any grid size
1 x ( x + 1) Formula for triangular
2 numbers
T= (Triangular number) n + Word Formula for any
(sum of triangular numbers but 1 below stair size)G stair size on any grid size
+ (Sum of triangular numbers but 1 below stair size)
T= ( 1 x(x + 1))n +(Σ triangular numbers less 1)G Improved word and algebraic
+ (Σ triangular numbers less 1) formula for any stair size
2 on any grid size
x-1 x-1
T= ( x ( x + 1) ) n + (Σ x (x+1) ) G + (Σ x (x+1) ) Algebraic formula for
2 x=1 2 x=1 any stair size on any grid
size
T= (x(x+1)) n + (Σ1(x2-1)) G + 1x(x2-1) An improved algebraic
2 6 6 formula for any grid
size on any stair size
