March 25 2011

Math IB SL

Internal Assessment – LASCAP’S Fraction

The goal of this task is to consider a set of fractions which are presented in a symmetrical, recurring sequence, and to find a general statement for the pattern.

The presented pattern is:

Row 1

1 1 Row 2

1 32 1 Row 3

1 64 64 1 Row 4

1 107 106 107 1 Row 5

1 1511 159 159 1511 1

Step 1: This pattern is known as Lascap’s Fractions. En(r) will be used to represent the values involved in the pattern. r represents the element number, starting at r=0, and n represents the row number starting at n=1. So for instance, E52=159, the second element on the fifth row. Additionally, N will represent the value of the numerator and D value of the denominator.

To begin with, it is clear that in order to obtain a general statement for the pattern, two different statements will be needed to combine to form one final statement. This means that there will be two different statements, one that illustrates the numerators and another the denominators, which will be come together to find the general statement. To start the initial pattern, the pattern is split into two different patterns; one demonstrating the numerators and another denominators.

Step 2: This pattern demonstrates the pattern of the numerators. It is clear that all of the numerators in the nth row are equal. For example all numerators in row 3 are 6. 1 1

3 3 3

6 6 6 6

10 10 10 10 10

15 15 15 15 15 15

Row number (n)| 1| 2| 3| 4| 5|

Numerator (N)| 1| 3| 6| 10| 15|

N(n+1) - Nn| N/A| 2| 3| 4| 5|

Table 1: The increasing value of the numerators in relations to the row number. From the table above, we can see that there is a downward pattern, in which the numerator increases proportionally as the row number increases. It can be found that the value of N(n+1) - Nn increases proportionally as the sequence continues.

The relationship between the row number and the numerator is graphically plotted and a quadratic fit determined, using loggerpro.

Figure 1: The equation of the quadratic fit is the relationship between the numerator and the row number. The equation for the fit is: N= 0.5n2+0.5n or n2+n2, n>0 Equation 1 In this equation, N refers to the numerator. Therefore, N= 0.5n2+0.5n or n2+n2, n>0 is a statement that represents step 2 and also step 1.

Step 3: In relation to table 1 and step 2, a pattern can be drawn. The difference between the numerators of two consecutive rows is one more than the difference between the previous numerators of two consecutive rows. This can be expressed in a formula N(n+1) - N(n) = N(n) - N(n-1) + 1. For instance, N(3+1) - N(3) = N(3) - N(2) +1. Using this method, numerator of 6th and 7th row can be determined. To find the 6th row’s value, n should be plugged in as 5 so that N(6) can be found. As for the 7th row’s numerator, n should be plugged in as 6. 6th row numerator is therefore: N(5+1) - N(5) = N(5) - N(4)+1 N(6) – 15 = 15 – 10+1

N(6) = 15+6

N(6) = 21

7th row numerator is therefore: N(6+1) - N(6) = N(6) - N(5)+1 N(7) – 21 = 21 – 15 +1

N(6) = 42 – 15 + 1

N(6) = 28

Not only by this method, but from the equation found in step 2, figure 1, 6th and 7th row numerator can be found also. 6th row numerator: N(6)=0.5×62+0.5×6 N(6)=0.5×36+3...

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