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Lacsap's fraction math portfolio

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Introduction

Lacsap's fraction

Ryohei Kimura

IB Math SL 1

Internal Assessment Type 1

Lacsap’ fraction

        Lacsap is backward word of Pascal. Thus, the Pascal’s triangle can be applied in this fraction.

How to find numerator

In this project, the relationship between the row number, n, the numerator, and the denominator of the pattern shown below.

1

1

1

image00.png

1

1

image01.png

image01.png

1

1

image02.png

image03.png

image02.png

1

1

image04.png

image05.png

image05.png

image04.png

1

Figure 1: The given symmetrical pattern

image57.png(Biwako)

Figure 2: The Pascal’s triangle shows the pattern ofimage06.png

.It is clear that the numerator of the pattern in Figure 1 is equal to the 3rd element of Pascal’ triangle which is when r = 2. Thus, the numerator in Figure 1 can be shown as,

(n+1)C2  image07.png

[Eq.1]

where n represents row numbers.

Sample Calculation

- When n=1        

(1+1)C2  image08.png

(2)C2  image09.png

-When n=2

(2+1)C2  image08.png

(3)C2  image10.png

-When n=5

(5+1)C2  image08.png

(6)C2 image11.png

15

...read more.

Middle

x = 28                                        [Eq. 3]

How to find denominator

                                      1   )+0

1   )+0

image12.png

image13.png

image13.png

image14.png

image15.png

image14.png

image16.png

image17.png

image17.png

image16.png

Figure 3: The pattern showing the difference of denominator and numerator for each fraction. The first element and the last element are cut off since it is known that all of them are to be 1. However, only first row is not cut off.

Table 1: The table showing the relationship between row number and difference of numerator and denominator for each 2nd element

Row Number (n)

Difference of Numerator and Denominator

1

0

2

1

3

2

4

3

5

4

The difference of numerator and denominator increases by one. Moreover, it is clear that the difference between row number and difference of numerator and denominator is 1. Thus, the difference can be stated as (n-1). Therefore, the denominator of the 1st element can be shown as,

image18.png

                                    [Eq. 4]

...read more.

Conclusion

th row can be solved as,

1st element

image25.png

image26.png

image27.png

2nd element

image28.png

image29.png

image30.png

13

3rd element

image31.png

image32.png

image30.png

12

4th element

image33.png

image34.png

image35.png

13

5th element

image36.png

image37.png

image38.png

Therefore, the pattern in 6th row is

image39.png

Also, the denominator in 7th row can be solved as,

1st element

image40.png

image41.png

image42.png

2nd element

image43.png

image44.png

image30.png

18

3rd element

image45.png

image46.png

image30.png

16

4th element

image47.png

image48.png

image35.png

16

5th element

image49.png

image50.png

image51.png

6th element

image52.png

image53.png

image54.png

Hence, the pattern in 7th row is,

image55.png

Conclusion

Therefore, the general statement of the rth element in nth row can be shows as,

image56.png

                                    [Eq.7]

where r is element number,  

However, there are several limitations for this equation. First, number 1 located in both side of the given pattern should be cut out when the numerator is calculated. Thus, the second element of each row is counted as “the first element.” Second, n in general statement of numerator must be greater than 0. Third, the very first row of the given pattern is counted as the 1st row.

...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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