This document describes a student's analysis of the pattern in "Lacsap's Fractions", which resembles Pascal's triangle. The student tests a theory that the numerators can be found using the formula xn = n(n+1)/2, where n is the row number. This formula is verified to work for sample rows. The student also finds patterns to determine the denominators using relationships to the row number and element position. These formulas are tested on sample calculations and used to determine the elements in the 6th row of the pattern.

1. Lacsap’s Fraction
Erika Mingo
IB Math SL 1
Internal Assessment Type 1
In this task “Lacsap’s Fractions” the following pattern occurs:

2. 1 1
3
1 2 1
6 6
1 4 4 1
10 10 10
1 7 6 7 1
15 15 15 15
1 11 9 9 11 1
Through research and previous knowledge, I realized that “Lacsap’s fractions” is very
similar to Pascal’s Triangle pictured below. The rule to solve the triangular numbers in
the pattern is x = n(n+1)/2, where n= the row number. I will test this theory to see if this
rule can be applied to Lacsap’s Fractions to find the numerator in the sixth and seventh
rows.
Finding the Numerator:
First I will test this rule, xn = n(n+1)/2, in a sample calculatation, where n= the row

3. number.
Row 1: 1 1
3
Row 2: 1 2 1
6 6
Row 3: 1 4 4 1
10 10 10
Row 4: 1 7 6 7 1
15 15 15 15
Row 5: 1 11 9 9 11 1
Sample Calculation
1. xn = n(n+1)/2
when n= 2
x2 = 2(2+1)/2
=2(3)/2
=6/2
=3
2. xn = n(n+1)/2
when n= 5
x5 = 5(5+1)/2
=5(6)/2
=30/2
=15
The equation xn = n(n+1)/2 can be applied to find the numerators of “Lacsap’s
Fraction”, and therefore can be used to find the 6th and 7th row’s numerators.

4. 6th Row:
xn = n(n+1)/2
when n= 6
x6 = 6(6+1)/2
=6(7)/2
=42/2
=21
Therefore, sixth row’s numerator is 21.
7th Row:
xn = n(n+1)/2
when n= 7
x6 = 7(7+1)/2
=7(8)/2
=56/2
=28
Therefore, the seventh row’s numerator is 28.
Table 1: This table shows the relationship between the row number and the value of the
numerator
Row Number (n) Numerator
1 1
2 3
3 6
4 10
5 15
6 21
7 28

5. Values of the Numerators
As They Relate to the Row Number
30
25
a lu e s
20
V
15
e r a t o r
10
u m
5
N
0
row 1 row 2 row 3 row 4 row 5 row 6 row 7
Row Numbers
This graph depicts the relationship between row numbers and the values of the
numerators
Finding the Denominator:
When looking at ‘Lacsap’s Fractions”, I found a pattern between the differences of the
denominator and numerator that is similar to Pascal’s Triangle.
This depicts the pattern of the differences of the denominator and the numerator.
Row 1: 1}0 1}0
3
Row 2: 1 2 }1 1
6 6
Row 3: 1 4 }2 4 }2 1
10 10 10
Row 4: 1 7 }3 6 }4 7 }3 1
15 15 15 15
Row 5: 1 11 } 4 9 }6 9 }6 11 }4 1
This depicts the pattern of the differences of the denominator and the numerator.

6. Table 2: This table shows the relationship between the row number and the 1st element in
each row. For the purpose of this part of the task, you must disregard the 1s along
the sides of the pattern.
Row Number Difference of Denominator
(n) and Numerator
1 0
2 1
3 2
4 3
5 4
As you can see in the table, the difference of the denominator and numerator is equal to
the row number +1. Therefore I can make the statement:
Denominator= Numerator - (n -1)
Sample Calculation
1.
3
2
= 3 - (2 - 1)
=3-1
=2
The denominator is 2.
10
2. 7
= 10 - (4 - 1)
= 10 - 3
=7
The denominator is 7.

7. Table 3: This table shows the relationship between the row number and the second
element in each row. For the purpose of this part of the task, you must disregard the
1s along the sides of the pattern.
Row Number Difference of Denominator
(n) and Numerator
1 N/A
2 N/A
3 2
4 4
5 6
The difference of the denominator and numerator for the 1st and 2nd rows are not
applicable because these rows only contain 1 element. A pattern that can be seen in this
table an increase of 2 as it descends down the rows. This means that the difference of the
denominator and numerator can be expressed as 2(n-2). Therefore you can find the
denominator solve the denominator by:
Denominator2= Numerator2 - 2(n-2)
Sample Calculation
10
1. 6
= 10 - 2(4-2)
= 10 - 2(2)
= 10 - 4
=6
The denominator is 6.
15
2. 9
= 15 - 2(5-2)
= 15 - 2(3)
= 15 - 6
=9
The denominator is 9.

8. Therefore, the denominator can be expressed as:
(let r represent the element number)
Denominator= n(n+1)/2- r(n - r)
Sample Calculation
15
1. 9 r= 2 n= 5
Denominator= 5(5+1)/2 - 2(5-2)
= 5(6)/2 - 2(3)
= 30/2 - 6
= 15 - 6
=9
The denominator is 9
10
2. 7 r= 3 n= 4
Denominator= 4(4+1)/2 - 3(4-3)
=4(5)/2 - 3(1)
=20/2 - 3
=10 - 3
=7
The denominator is 7.
Therefore, the denominators in the 6th row can be solved:

9. 1st Element Denominator1= 6(6+1)/2- 1(6 - 1)
=6(7)/2 -1(5)
=42/2 - 5
= 21- 5
= 16
nd
2 Element Denominator2= 6(6+1)/2- 2(6 - 2)
=6(7)/2 -2(4)
=42/2 - 8
= 21- 8
= 13
rd
3 Element Denominator3= 6(6+1)/2- 3(6 - 3)
=6(7)/2 -3(3)
=42/2 - 9
= 21- 9
= 12
th
4 Element Denominator4= 6(6+1)/2- 4(6 - 4)
=6(7)/2 -4(2)
=42/2 - 8
= 21- 8
= 13
th
5 Element Denominator5= 6(6+1)/2- 5(6 - 5)
=6(7)/2 -5(1)
=42/2 - 5
= 21- 5
= 16
21 21 21 21 21
Therefore, the pattern for the 6th row is: 16 , 13 , 12 , 13 , 16