1. Emmanuel Castaño
Ms. Bessette; Period 7
Due: March 30th, 2011
IB SL1 Math Internal Assessment – Lacsap’s Fractions
This assignment required students to find patterns within a triangle of elements which
had noticeable similarities to Pascal‟s triangle; a triangle of number‟s commonly used for
binomial expansions. An example of Pascal‟s triangle can be seen in Figure 1 below:
Row 1
Row 2
Row 3
Row 4
Row 5
Figure 1: These are the first five rows of Pascal’s triangle.
The variable is the number of the element in a row, or the number of the column the
element is in. The first element of each row is , and thus is the of each row. The
equation to find a particular element , when is the row number, and is a combination from
probability, is:

2. The triangle researched for this internal assessment had the same format as Pascal‟s
triangle; however it included fractions and more complex patterns between elements. The first
five rows of the triangle were given, and the students were required to find the elements of the
next two rows. The five rows given were:
Figure 2: These are the first five rows of Lacsap’s Fractions.
From this example it can be observed that the numerator of each row remains the same
throughout the whole row, supposing that the first and last term in a row is a fraction. In which
case the first and last terms in the second row two would be , in the third row, and the other rows
would follow this pattern successively. A pattern that can be used to find the numerator of the sixth and
seventh row is shown below:
Row 1
Row 2
+3
Row 3
+4
Row 4
+5
Row 5
Figure 3: This is a simple pattern in numerators of the first five rows of Lacsap’s Fractions.
This pattern shows that the numerator values are growing exponentially in comparison to
the row numbers of the numerators.

3. This simple pattern can be used to predict that the next to numerators are 21, and 28
respectively. However, to find an equation that could calculate the numerator of the numerator in
the row, a pattern had to be noticed between the row number and the numerator of the row,
instead of the between the numerator from previous rows and the numerator of row .
To find an equation that could predict the numerator of any row, I set up three systems of
equations and used the method, substitution, to solve for the coefficients.
Row Number Numerator
1 1
2 3
3 6
4 10
5 15
6 21
7 28
Table 1: is a table showing
numerators of each row of
Lacsap’s Fractions.
As has been previously stated, the relationship between and the numerator is exponential, and
therefore, the equation for these two sets of values must be a quadratic function. Therefore:
When the values for in Table
1 are entered:
When the values for in Table
1 are entered:
Or:
Or:

4. By using substitution:
Or:
This can be simplified to:
If this and are then substituted
into , then:
Or:
This can be simplified to:
If this is plugged back into
=− − +1, then:
Or:
Or:
From there, it can be plugged into
If then the values for are =2−3 , and which returns,
entered as a quadratic:
, and respectively.
The solution to this system of equations return the values , , and . This
results in the quadratic equation, , or . When is replaced by the for
the row number, and is replaced by the numerator:
This equation states that the numerator of any row will be the row number multiplied
by one more than itself, divided by two.

5. The table below shows how the numerator of a row can be found with by using
.
Row Number Equation Numerator
.
1 . 1
2 . 3
3 . 6
4 . 10
5 . 15
6 . 21
7 . 28
Table 2: is a table showing how the numerator
of each row can be reached from the row
number.
Figure 4: shows how the equation goes through the values in “Table 2.”

6. The denominator in this triangle is more complicated to predict since it changes between
rows, and in each position within each row. To complete the denominators for the sixth and
seventh row I used the simple pattern shown in Figure 5. By following the pattern shown below,
the denominator can be deduced to be 16, 13, 12, 13, and 16 respectively for equals 1, 2, 3, 4,
and 5 in the sixth row, and 22, 18, 16, 16, 18, and 22 for equals 1, 2, 3, 4, 5, and 6 in the
seventh row.
Row 1
Row 2
+2
Row 3
+3 +2
Row 4 +2
+4 +3
Row 5
Figure 5: This is a simple pattern in the denominators of the first five rows of Lacsap’s Fractions.
By noticing the pattern shown in Figure 5 and by using for the
numerator, I found the sixth and seventh row of the triangle:
Figure 6: These are the first seven rows of Lacsap’s Fractions.

7. Even though the pattern used to find the sixth and seventh row of the triangle is very
simple, to find a general statement for the denominator of the term of the row, a pattern
has to be found within each row in terms of . Even though there are many patterns that can be
noticed such as, the middle terms of odd rows can be found by adding one to the row number,
squaring it, and dividing it by two, , a useful pattern must work for all rows and all
elements.
I noticed that the numerators are always greater than the denominators which led me to
experimenting with subtracting from the numerator. However, since the denominator decreases
until the middle term, and continues to increase symmetrically as it reaches the end of a row, the
number subtracted from the denominator had to have or to have the subtracted amount be
the same on either side of the middle term in a row.
Then I realized that the pattern for the denominator was quadratic. It started high,
lowered to a vertex, and then increased to its starting point. I also noticed that if the first and last
term of each row follow the pattern that the numerator is the same throughout the whole row,
then to equal, „1,‟ the first and last term must be the numerator over itself. Therefore, when
the denominator is the same as the numerator, and therefore, y-intercept of the quadratic
equation must be the numerator.
To find an equation that could predict the denominator of any element of any row, I set
up three systems of equations and used the method, substitution, to solve for the coefficients.

8. Element Denominator Row
0 1 3 6 10 15 21 28
1 1 2 4 7 11 16 22
2 3 4 6 9 13 18
3 6 7 9 12 16
4 10 11 13 16
5 15 16 18
6 21 22
7 28
Table 3: is a table showing how the relationship between the number of the element
and the denominators in the first seven rows of the triangle.
As has been previously stated, the relationship between and the numerator is a
quadratic relationship, and therefore, the equation for these two sets of values must be a
quadratic function. Therefore:
By using substitution:
When the values for for the 3rd
row in Table 3 are entered: This can be simplified to:
Or: Or:
When the values for in Table If this is plugged back into
3 are entered: =− − + , then:
Or: Or:

9. If then the values for in Table This can be simplified to:
3 are entered as a quadratic:
Or:
Or:
From there, it can be plugged into
If this and are then substituted , and which returns,
into , then: and respectively.
The solution to this system of equations return the values , , and . This
results in the quadratic equation, , or . When is replaced by
the for the number of the element in the row, and is replaced by the value of the element:
As predicted, was equal to the numerator of the . However, with only one row
being tested, a pattern has not being found for the coefficients and . The same method will be
used to solve a system of equations for the fourth row of the triangle.
Or:
When the values for for the 4th
row of the triangle are entered:
By using substitution:
Or: This can be simplified to:
When the values for in the Or:
triangle are entered:

10. If this is plugged back into If this and are then substituted
=− − + , then: into , then:
Or: This can be simplified to:
If the values for in the triangle Or:
are entered as a quadratic:
From there, it can be plugged into
Or: , and which returns,
and respectively.
The solution to this system of equations return the values , , and . This
results in the quadratic equation, , or . When is replaced
by the for the number of the element in the row, and is replaced by the value of the
element:
In the 3rd row, the equation to find the values in the denominator was, , and
in the 4th row, the equation was . In both of these equations, the value was, „1,‟
the value was equal to the negative row number , and the value was equal to the
numerator of that row. By following this pattern, the equation for the denominator of the
element of the row would be:
Where is the numerator, or .

11. Element Denominator Row
0 1 3 6 10 15 21 28
1 1 2 4 7 11 16 22
2 3 4 6 9 13 18
3 6 7 9 12 16
4 10 11 13 16
5 15 16 18
6 21 22
7 28
Table 4: is a table showing how the relationship between the number of the element
and the denominators in the first seven rows of the triangle.
Table 4 shows the values that will be graphed below. If the equation
is accurate, the graphs, , , , ,
, , , and , should pass through the points of the
corresponding row.

12. Figure 7: shows the relationship between the number of the element and the denominator of
the first seven rows.

13. The quadratic equations shown in Figure 1 show that for each one of the first seven rows:
When is replaced with and is replaced with :
is also the numerator of the row, or :
can be factored out:
Therefore, if is added to the numerator, of any row , the denominator of term
can be found. If the equation for the numerator, , and the equation for the denominator,
, are combined, it would yield an equation to find any element on any row of
the triangle.
By using this equation, any term from the triangle below can be found. In the case that
or , the part of the general equation, will equal to and therefore, the
equation would simplify to, . Therefore the only limitation to this equation is
that the first and last term of each row will equal the numerator over itself. Even though this
simplifies to „ ,‟ some of the terms in the triangle, such as which equals , are not
simplified. With this in mind, before entering a value into the general equation above, if or
, then,
The scope of this equation is for whenever , and , when and can only be
integers. This is because the triangle‟s first element is element 0, and its first row is row 1.

14. The following values were acquired by using the formula .
Row Number Element Number Equation Element
1 0 .
5 2 .
7 5 .
8 8 .
Table 5: is a table showing how the values in “Figure 8”were found.
Figure 8: These are the first ten rows of Lacsap’s Fractions as would be found with
.

15. “I, the undersigned, hereby declare that the following assignment is all my own work and that I
worked independently on it.”
“In this assignment, I used LoggerPro 3.8.2 to draw my graphs.”