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Lascap's Fractions. I was able to derive a general statement for both the numerators and denominators, and prove it for other rows of the pattern

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Introduction

Elisabetta Sorrentino

Lacsap’s Fractions

Math SL Internal  Assessment TYPE I

Name: Elisabetta Sorrentino

Subject: Math

Candidate Number: 001459 - 048

School: Suzhou Singapore International School

Pages: 14

Lacsap’s Fractions

Aim: In this task I will consider a set of numbers that are presented in a symmetrical pattern. I will attempt to find the next 2 rows of the pattern and a general statement to find any term. Further more, I will also examine the validity, scope and limitations of my general statement.

image00.png

As shown on the diagram, the numerator of the fraction increases as we go down the rows. The pattern is straightforward; it is an addition of the row number to the numerator of the previous row.  The amount added is increased by one each row. This pattern can be easily understood in the table below.

Row Number (n)

Numerator = (Numerator of the previous row + n)

1

1      (0+1)

2

3      (1+2)

3

6      (3+3)

4

          10    (6+4)

5

15    (10+5)

6

21    (15+6)

Both the numerator and denominator have a consistent pattern when the 1’s from the first row are not considered. Since the first row is not considered, the pattern consequently starts from row 2.

Using Microsoft Excel, I plotted the relation between the row number, n, and the numerator in each row.

From the graph above, I noticed that the trend line of the relation between the row number and the numerator in each row is quadratic.

...read more.

Middle

Denominator

11

9

9

11

Consequently, using Microsoft Excel, I plotted the relationship between the element number and the denominator. The graph below shows this relationship.

To represent this graph in an equation for the denominator, I used a graphic calculator to derive the equation using the quadratic regression function, following these steps.

  1. Click STAT, then click “1:Edit”

image01.png

  1. Put the element in L1and the denominators for row 5 on L2image02.png
  1. Then click STAT, select CALC, then select “5:QuadReg”image03.png
  1. Click 2ND, input L1, then comma, input L2

image04.png

  1. Then comma again, click VARS, then select Y-VARS, select “1:Function…” image05.png
  1. Then click “1:Y1

image06.png

  1. The answer got is: image07.png

Let n be the row number, D the denominator, r the element number and y the numerator in the row (changing the x,y,b,c variables)

As we can see from step 14 above, b is negative and therefore when we substitute the variable n in the equation it is also negative. Additionally, the numerator, y, is substituted for the variable c, as it represent the numerator of the row.  

The working below proves the equation right for the second denominator element in the fifth row.

The following working proves the equation right for the first denominator element in the fourth row.

...read more.

Conclusion

Secondly, the variable n, must be greater than 0. This because there cannot be negative rows, as if plugged into the general statement the answer would be positive, when it should be negative in order to be correct.

Thirdly, the variable r, must be equal or greater than 0, as there cannot be 0 or negative elements in a pattern. For the equation to work, the variable r, has to be an integer, fractions will not work.

Lastly, the element number, r, must be smaller then the row number n. As we can see from the table below the element is always one less then the row number. Therefore the element, r, cannot be greater then the row number, n, as a row number can only have a limited amount of elements.

Row number

5

6

7

8

9

10

11

Element

4

5

6

7

8

9

10

In conclusion, through the use of a straightforward pattern, I was able to find the next 2 rows of the given symmetrical pattern. In addition with the use of quadratic regression I was able to derive a general statement for both the numerators and denominators, and prove it for other rows of the pattern. With the help of these two equations, I finally arrived at my general statement. Lastly, thanks to the deeper understanding of the calculation process, I examined in detail the validity, scope and limitations of my general statement.

...read more.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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