Knowing this, in the equation nCr, r would have =2 in this case. Afterwards, I decided to plot the row numbers of LACSAP’s fractions with the numerators and the newfound equation nC2, or n!/2!(n2)!
Task 2  Plot relation between the row number and Numerator, write general statement about it
This is a graph of Numerator (y – axis) vs Row Numbers (n) (x – axis)
In the equation nCr, n must be equal to or greater than r in order for it to work properly. Though the equation was coming up with the right numerators, they were coming in an order where it would be n+1, as opposed to n by itself. So I decided to try the formula (n+1)C2 instead to see what the results would be.
The values of (n+1)C2 came out to be the same as the numerator values, so I decided this would be more formula to find my other numerators.
Numerator = (n+1)C2
Task 3  Finding the 6th an 7th Numerators
6th Numerator
(6+1)C2
(7)C2
6!/2!(6 – 2)!
6!/2!4! = 21
7th Numerator
(7+1)C2
(8)C2
8!/2!(8– 2)!
8!/2!6! = 28
These numbers can also be found going down r = 2 of Pascal’s triangle.
Finding the Denominator
Whilst it took a bit of trial of error, I found that the difference between the Numerator and the Denominator were following a pattern similar to Pascal’s Triangle, eg. In this grid, I have replaced the numbers with the differences between the Numerator and Denominator to clearly see the pattern.
Comparing the new triangle to Pascal’s Triangle, it almost appears that the new triangle is a mirror image of Pascal’s triangle. Using this new information, I decided to plot the row numbers of LACSAP’s triangle against r = 1 of the new triangle as it’s the only diagonal row that appears consistent with Pascal’s triangle.
According to the table, the difference of the Numerator and Denominator is (n 1) as the values for the differences, though the same, are only the same once they’ve been bumped down one.
With this, I came up with the equation that D = N  (n – 1) where D is the Denominator to be found and N is the numerator for it to be found for.
Comparing the results of D = N  (n – 1), the results did not match up with the given denominators.
I decided to compare the second element, or r = 2 of the new element instead to see what values I would yield instead.
Looking at this table, the differences between the values for the difference between the Numerator and Denominator increase by 2. The formula for the relationship between the row number (n) and the Difference of the Numerator and Denominator came out to 2(n – 2), so I tried D = N  2(n – 2) instead.
Again, the values were different from the given Denominators. However, I noticed that when I was dealing with the 1st element and the 2nd element in the formula, they were correspondent to the numbers that kept changing between the formulas. Knowing this, I came up with a new equation for the Denominator, which is:
D = N  r(n – r) where r is the element number, n is the row number and N is the numerator.
Given that the Numerator formula is (n+1)C2, this equation can be rewritten out as:
Denominator =( (n+1)C2)  r(n  r)
Thus, the sixth row would be:
1, 21/16, 21/13, 21/12, 21/13, 21/16, 1
The seventh row would be:
1, 28/22, 28/18, 28/16, 28/16, 28/18, 28/22, 1
Task 4 – Find the General Statement for En(r)
Now that I know how to find the Numerator and the Denominator, the equation to find a fraction in LACSAP’s triangle, En(r), can be written as:
Numerator/Denominator = (n+1)C2/( (n+1)C2)  r(n  r)
Task 5 – Find Additional Rows
Row 8
(8+1)C2/( (8+1)C2)  r(8  r)
After this, the rest of the row is reflected, thus it looks like
1, 36/29, 36/24, 36/21, 36/20, 36/21, 36/24, 36/29, 1
Row 9
(9+1)C2/( (9+1)C2)  r(9  r)
1, 55/47, 55/41, 55/37, 55/35, 55/35, 55/37, 55/41, 55/47, 1
Row 10
(10+1)C2/( (10+1)C2)  r(10  r)
1, 55/46, 55/39, 55/34, 55/31, 55/20, 55/31, 55/34, 55/39, 55/46, 1
Task 6  Discuss the scope/limitations of the General Statement
Some limitations of General Statement are that:
 The numerator must be greater than 0

In the equation (n+1)C2, n +1 must be greater than 2/r in order for the General Statement to work