- Level: International Baccalaureate
- Subject: Maths
- Word count: 2182
MATH Lacsap's Fractions IA
Extracts from this document...
Introduction
Mathematics SL
Internal Assessment Type I
By Claudia Cheng
Introduction
In this task, the goal is to consider a set of numbers which are presented in a recurring, symmetrical pattern and to find the general statement for En(r) be that the (r+1) th element is in the nth row and starting with r=0.
Figure 1: Lacsap’s Fractions
At first glance, I quickly noticed that the word “Lacsap” is “Pascal” spelt backwards. Therefore, the numbers presented in Figure 1 has a strong resemblance to Pascal’s Triangle.
Figure 2: Pascal’s Triangle[1]
Finding the general statement for numerators
Upon observing Pascal’s Triangle, I noticed that the numbers on the third diagonal row, as highlighted with a blue box, in Figure 2 are the same as the numerators in Lacsap’s fractions in Figure 1. For example, the first number is 1, the second is 3, the third is 6, the fourth is 10 and the fifth is 15. These set of numbers are known as triangle numbers. Triangle numbers enumerate the items that can be arranged in a triangle. Thus, this is useful in founding the patterns in Lacsap’s Fractions. Looking at Pascal’s Triangle, we can then assume the numerator for the sixth row to be 21.
Middle
Denominator = Numerator – 3(n – 3)
Looking at the three equations derived above, there is an emergence of a pattern. The numbers in each equation match up to their element number hence they are substitutable. Using the format of the first equation deduced from element 1, the element number (r) could be substituted for 1. This results a general statement for the nth term of the denominator where D is the denominator and N is the numerator:
D = N –r(n – r)
or
The general statement for En(r)
We have now obtained the general formula for both the numerator and denominator values. Both these formulas can be combined to form a general statement.
=
Finding the sixth row
The general statement will firstly be used to find the sixth and seventh rows. To find the sixth row, the row number will be inputted for n and the element will also be inputted for r. The working is shown in Table 6.
Table 6: Calculating the values on the sixth row
E6 (1) = |
E6 (2) = |
E6 (3) = |
E6 (4) = |
E6 (5) = |
As mentioned before, the “1’s” were discarded whilst doing the investigation, but now it must be added back into the beginning and end of the row.
Conclusion
Conclusion
I have concluded with my general statement through a process. Firstly, I found the general statement for the numerators in each row through observations and calculation through simultaneous equations. This was then validated through the use of Microsoft Excel. To find the pattern for the denominator, I noticed a relationship between the numerator and the denominator. By splitting the elements and using the differences as a base, I noticed that the general statement for the denominator also had a correlation with the element number and row numbers. This allowed me to calculate a general statement for the denominator pattern. With both the statement for the numerator and the statement for the denominator, I was able to combine them to form the general statement, En (r).
Figure 4 is a newly devised figure of Lacsap’s fractions that has all the values up to the 9th row Through Pascal’s triangle numbers, the use of graphing technology, the observation of symmetry for the denominator, I have drawn to the conclusion that my statement is valid for all the values I have calculated and tested.
[1] Image taken from <http://www.janak.org/pascal/pascal1.html>, November 12th 2012
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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