Portfolio #1
This portfolio surrounds the mathematical ideals of the LACSAP’s Fractions, and creating the task of answering certain questions about a specific symmetrical pattern. Through a work entirely of my own and without any unauthorized outside assistance, I answers all of the questions in this portfolio along with showcasing all my work, aided by the use of technology and patterns discovered by me.

The symmetrical pattern provided possesses only 5 vertical rows, with number of elements r increasing by 1 per new row created, and with r=0 representing the first element on each. However, one of the tasks requires finding the 6th row and its elements through patterns found in the pattern itself. Through a close analysis of the symmetrical pattern, which resembles Pascal’s Triangle in many ways, I found a relation between the numerators of the 1st element of every row, with r=1. As the first element of row 1 equals 1 (which can also be written as 11), the first element of row 2 equals 32. Just through that it can be seen the difference between numerators equaling 2. And as the first element in the third row is 64, the difference between numerators of second and third rows equals 3. And as I continued to analyze the numerators on the any elements of each of the following two rows, I came to the conclusion there was a pattern between their numerators and their row numbers. Therefore, the numerator of any element in any row will result from next numerator=previous numerator+(previous difference+1) equation. To be more mathematical, I developed the equation of numerator(row n)=n2+n2 , with n equaling the row number. To validate my general statement for finding numerators for rows, I tested it for finding the 6th row’s numerator, common to all of its elements. I calculated numerator6th row=n2+n2= 62+62=36+62=422=21, and through re-checking the patterns I’d found earlier and applying it to that row I came to the same results. However, the task was not fully...

...Lacsap’sFractions
IB Math 20 Portfolio
By: Lorenzo Ravani
Lacsap’sFractions Lacsap is backward for Pascal. If we use Pascal’s triangle we can identify patterns in Lacsap’sfractions. The goal of this portfolio is to ﬁnd an equation that describes the pattern presented in Lacsap’sfraction. This equation must determine the numerator and the denominator for every row possible.
Numerator
Elements of the Pascal’s triangle form multiple horizontal rows (n) and diagonal rows (r). The elements of the ﬁrst diagonal row (r = 1) are a linear function of the row number n. For every other row, each element is a parabolic function of n. Where r represents the element number and n represents the row number. The row numbers that represents the same sets of numbers as the numerators in Lacsap’s triangle, are the second row (r = 2) and the seventh row (r = 7). These rows are respectively the third element in the triangle, and equal to each other because the triangle is symmetrical. In this portfolio we will formulate an equation for only these two rows to ﬁnd Lacsap’s pattern. The equation for the numerator of the second and seventh row can be represented by the equation: (1/2)n * (n+1) = Nn (r) When n represents the row number. And Nn(r) represents the numerator Therefore the numerator of the sixth...

...Exploration of Lacsap’sFractions
The following will be an investigation of Lacsap’sFractions, that is, a set of numbers that are presented in a symmetrical pattern. It is an interesting point that ‘Lacsap’ is ‘Pascal’ backwards, which hints that the triangle below will be similar to “Pascal’s Triangle”.
1 1
1 1
1 1
1 1
1 1
There are many patterns evident in this triangle, for instance I can see that there is a vertical axis of symmetry down the middle of the triangle. Each row starts and ends with the number 1. Each row has one more variable than the number of rows, i.e. row 1 has 2 variables. The numerators in the middle stay the same and the diagonals form sequences.
In order to decipher the pattern in the numerators and denominators, I had to look at the triangle a different way. Knowing that the numerators of the row don’t change, it occurred to me that the number 1s on the outside of the triangle could be expressed as fractions.
This proves that all the numerators of the row are the same.
To further investigate the numerators, I will examine the relationship between the
row number and the numerator, which is shown in the table below. These are the numerators after having...

...Investigation
Lacsap’sFractions
The focus of this investigation is surrounding Lascap’s Fractions. They are a group of numbers set up in a certain pattern. A similar mathematical example to Lacsap’sFractions is Pascal’s Triangle. Pascal’s Triangle represents the coefficients of the binomial expansion of quadratic equations. It is arranged in such a way that the number underneath the two numbers above it, is the sum.
Ex. 1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
In the example of Pascal’s Triangle below, the highlighted numbers represent this pattern. The two numbers above the third add up to equal the third. (e.g. 2+1=3)
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
Although Pascal’s Triangle is set up similarly to Lacsap’sFractions, the patterns are different and the numbers are set up in fractions as opposed to whole numbers. The first five rows of Lacsap’sFractions are set up like this:
1
1 3/2 1
1 6/4 6/4 1
1 10/7 10/6 10/7 1
1 15/11 15/9 15/9 15/11 1
In the next part of the investigation, we will explore the patterns...

...Math SL I.A:
Lacsap'sFractions
Introduction
In my internal assessment, type 1, I was given Lacsap'sFractions task. To do my calculations I used a TI-84 graphic calculator. To type the I.A I used Apple's Pages, Microsoft Excel 2011 and Microsoft Word 2011.
Lacsap'sFractions
To find the numerator of the sixth row I looked at the difference between each of the numerators.
1 1
1 32 1
1 64 64 1
1 107 106 107 1
-------------------------------------------------
1 1511 159 159 1511 1
Row | Numerator | Difference between numerators n-nn-1 |
1 | 1 | 1 |
2 | 3 | 2 |
3 | 6 | 3 |
4 | 10 | 4 |
5 | 15 | 5 |
6 | 21 | 6 |
7 | 28 | 7 |
From this table one can notice that the difference between each numerator on each row is always d+1, where d represents the difference between the two previous rows. The equation representing this is: un=un-1+(d+1), where un represents the numerator you are looking for, n is the row number and un-1 is the numerator from the previous row.
Therefore to find the numerator of the 6th row I did:
u6=15+5+1
u6=21
After that, I plotted a graph doing numerator vs. row:
The relationship between the numerator and the row is best described by the equation of the line:
y=0.5x2+0.5x
The equation is quadratic and can be used to determine any numerator at any row. In the equation, x stands for the row and y represents...

...Lacsap’sFractions
The aim if this IA is to investigate Lacsap’sFractions and to come up with a general statement for finding the terms.
When I noticed that Lacsap was Pascal spelt backwards I decided to look for a connection with Pascal’s triangle.
Pascal’s triangle is used to show the numbers of ‘n’ choose ‘r’(nCr). The row number represents the value of ‘and the column number represents the ‘r’ value. Eg. Row 3, colomn 2 = 3C2 = 2.
I noticed that all the numerators of the fractions in Lascap’s fraction (3,6,10,15) are also found in Pascal’s triangle. So I tried to see if I would get the denominator of the fractions by using the row as ‘n’ and the colomn (or element) as ‘r’. This did not work out because Lascap’s triangle does not have a row with only one element like Pascal’s does. To solve this I just added 1 to each row number. This gives me the formula[pic].
|(Row number +1)C2 |Numerator |
|(2+1)C2 |= 3 |
|(3+1) C2 |= 6 |
|(4+1)C2 |=10 |
|(5+1)C2 |=15 |
Now that we have found an equation to solve to...

...INTRODUCTION
Lacsap’s triangle-The set of numbers in concern are basically an inverse of the Pascal’s triangle. These terms themselves are fractions which follow different series themselves. There is a specific function that can accurately predict the fractional numbers accurately. Using the graph plots we can calculate this function and predict the numbers accurately. The whole process for finding the adequate function would involve the use of different smaller function and thus create a general function.
The numerator follows a different equation and the denominator a different one.
1 1
1 1
1 1
1 1
1 1
Therefore if we only consider the numerators in the triangle
1 1
1 1
1 1
1 1
1 1
One can observe that the numerator values increase in a specific order.
So, Let r = number of rows; N= numerator.
Since the first row does not contain a fraction we keep its value as 1; we start from the second row onwards since it has a fraction.
Difference between two r values –
Nr3 – Nr2 = 6-3= 3
Nr4 – Nr3 = 10-6=4
Nr5 – Nr4 =15-10=5
As you can observe that the differences highlighted in yellow, increase by 1 as we the r increases.
Using this knowledge we could also find...

...1 1
1 32 1
1 64 64 1
1 107 106 107 1
1 1511 159 159 1511 1
The aim of this task is to find the general statement for En(r). Let En(r) be the element in the nth row, starting with r = 0.
First to find the numerator of the sixth row, the pattern for the numerator for the first five rows is observed. Since the numerator is the same in each row (not counting the first and the last number in each row), I can observe the numerator in the middle of each row. The numerators from row 1 to row 5 are 1,3,6,10,15
Table 1: A table showing the relationship between the row number and the numerator. The table also shows the relationship between the numerators in each row.
Row | Numerator | 1st differences | 2nd differences |
1 | 1 | 2 | 1 |
2 | 3 | | |
| | 3 | |
3 | 6 | | 1 |
| | 4 | |
4 | 10 | | 1 |
| | 5 | |
5 | 15 | | |
The difference between the numerator in row 1 and row 2 is 2, row 2 and row 3 is 3, row 3 and 4 is 4 and row 4 and 5 is 5. The second difference for each row number is 1; this shows that the equation for the numerator is a geometric sequence. So I try to find the equation of the sequence by using the quadratic formula, y = ax2 + bx + c, where y = the numerator and x = the row number.
6 = a(3)2 + b(3) +0
6 = 9a +3b
6 = 9a + 3(-2a + 1.5)
6 = 9a – 6a + 4.5...

...
LACSAP’SFraction-‐
Portfolio
Type
I
LACSAP’SFractions - Math SL Type I
Name: Yao Cia Hua
Date: March 22nd, 2012
Teacher: Mr. Mark Bethune
School: Sinarmas World Academy
1
Yao
Cia
Hua
Mathematics
SL
LACSAP’SFraction-‐
Portfolio
Type
I
Lacsap triangle is a reversed Pascal triangle. This task focuses mainly on finding the
relationship between the number of row n and the numerator N and also the
relationship between the element of a row r and the denominator D . Through this,
a general statement base for En (r) on N and D are suppose to be stated and
explained. For this task various technologies, such as Geogebra, MathType and
calculator, are needed in order to produce a more organized piece of work and
clearer graphs and diagrams.
Finding The Numerator In The Sixth Row:
In order to find the sixth row of this of this triangle, a pattern must firstly be found.
As it is seen from the diagram on the above, the pattern shows that, by adding 1 to
each of the difference between the 2 previous numerators, this will equate the
numerator of the next row.
For example: (using the first 2 numerators)
Difference: 3 – 1 = 2
The following numerator: 2 + 1 = 3
2
...