Type I Internal Assessment (Lascap's Triangle)

For example,

3C2 = (3!) / [( 2!) x (3 - 2)!] = 3

This can be used to find any number on the Pascal’s Triangle considering that the 1 on top of the triangle is represented by n = 0 and r = 0 and the 1 must not be counted as a number when looking for the ‘r’ value.

Therefore this can be used to find the numerators in the Lacsap’s Triangle.

Finding and plotting the relation between the row number and the numerator in each row and finding a general formula

I have made a table showing the relationship between the numerator and the row number.

The table shows that the difference between each subsequent numerator increases by 1 each turn starting with 2. We see that the first difference between the numerator in row 1 and 2 is 2 and this logic applies for the other rows as well. We also see that the second difference in all of them is 1.

Below you can see a graph showing the relationship between the numerator and the row number.

Even though we see the numerator and the row number rise at a different rate, they both contribute to the ascending line. We observe that the line looks like a half parabola. This observation once again suggests that the general statement for Lacsap’s triangle is a quadratic equation: ax2 + bx + c

To find a general formula for the numerator, I first tried to find the relationship between the row number (n) and the numerator (Sn).

I tried  

 so for the first row, it is. For row 2, it is

. I tried to find a relationship again using row 2 and found

  for row 2.

Therefore,

 and so

This can be rewritten as,

I used a graphic calculator to validate the general formula using a GDC and the QuadReg function.

Rewritten,                      or

Validating the formula manually

We can validate the formula by trying it on the numbers that we already know.

Finding the 6th and 7th rows

We can use the general formula to find the numerators of the 6th and 7th row.

6th row:

7th row:

Finding the denominator

After taking a hard look at the Lacsap’s Fractions, brain storming through all the possibilities and then through some trial and error, I found that the denominator too is related to the Pascal’s triangle. I have made an image which shows a Pascal’s triangle. In the image, I have highlighted the numbers which represent the difference in the numerator and the denominator of the Lacsap’s fractions.

We also observe that the difference in the numerator and the denominator if we go diagonally for the first element number is 1, 2,3,4,5 respectively.

Now using the numerator and ‘x’ as the difference between the numerator and denominator, we get the equation

The difference of the denominator for element 1 increases by one for every consecutive row. Considering this, we can say

In which n is the row number.

For the second element number, the difference between the numerator and denominator is 2, 4, and 6. Therefore the difference between the numerator and denominator for every consecutive number in the second number is 2. With this in mind, we can conclude

Similarly, for the third element number, the difference is 3. We can see a pattern emerging and can say that parts of the equation are interchangeable. Thus we can replace these numbers by the element numbers. And we get the equation for the denominator as:

                                                                               Where ‘r’ is the element number

OR

Denominators for the sixth row by using the equation,

1, 16, 13, 12, 13, 16, 1

Row 6 =

Denominators for the 7th row by using this equation,

1, 21, 18, 16, 16, 18, 21, 1

Row 7 =

General Formula

Since we know the two formulas, we can easily combine them to get a general formula.

Numerator =

Denominator =

Therefore, General Formula =

 where ‘n’ is the row number and ‘r’ is the element number.

I will now use this formula to find the 8th, 9th and the 10th rows.

Row 8:

 … 

Therefore Row 8 =

Row 9:

 … 

Therefore Row 9 =

Row 10:

 … 

Therefore Row 10 =

Using this general formula, elements from the Lacsap’s triangle may be found simply by substituting ‘n’ by the row number and ‘r’ by the element number in the triangle. But using this formula, you have to discard the 1 at the beginning and the end of each row. Thus it must be added again manually every time you find the elements of a row. This is one of the limitations of the formula. Another limitation of the general formula is that it cannot be used to find the elements in row 1 since it only has one element in it which is 1 and since the 1 was discarded from all rows, it cannot be used. Also because the 1 is removed, every element in the row is reduced by 2 due to which the element number starts from 1 and not 0. Also the element number has to be more than or equal to 2 and the row number has to be more than or equal to 1 for the general formula to work.

Conclusion

I used mathematics and a few observations to find the different formulas required to find the final general formula. I have also used modern technology to validate the formula of my numerator. This was done using my GDC – QuadReg function to validate the formula before moving on to finding the denominator. Using some observation and trial and error, I was able to find a relationship between the numerator and the denominator which in turn allowed me to find a general formula for the denominator. Finally, after finding these two formulas, I combined them in a simple way to get the general formula and used this formula to find additional rows to prove the formula works.