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Lacsap's Fraction Math Portfolio

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Introduction

Lacsap’s Fraction

Thomas Ullrich

Internal Assessment Type 1

Math SL

Introduction:

In this IA Type 1 I’ll have to find patterns in a set of numbers, which are presented in a symmetrical pattern.

Step 1: Finding the numerator of the 6th row

At first, I had a look at the first 4 numerator, which are 3, 6, 10, 15. The first step between 6 and 3 is 3 and increases by 1 in each following row. Therefore we can say that the numerator in the 6th row is 21 (and the 7th is 28.)

Figure 1 Relation between the numerator and n

As you can see the numerator develops not linearly, but in some way exponentially.

image00.png

Figure 2 Lacsap's Triangle

Step 2:

The next step is about finding the complete 6th and 7th row.  For the numerator, I found this specific pattern that describes it. As we see in our first graph, the numerator in row number 6 is 21 and in row number 7 it is 28.

Now it is about finding a pattern for the denominator.image01.png

To show the specific pattern for the denominator I marked the sequences which are responsible for the characteristical features of the Pascal’s triangle.

The sequences are:

2  (+2)  4 (+3)  7 (+4)  11

4  (+2)  6 (+3)  9

7  (+2)  9

Figure 4 Denominator pattern

As we can see in Figure 3, the denominator increases by 1 starting with a step of 2. Following this rule, the denominators are:

6th

...read more.

Middle

To find b, we put a in equation IV:

5=9x0.5+b

5=4.5+b

b=0.5

To find our last unknown variable c, we put a and b in I:

10=16x0.5+4x0.5+c

10=10+c

c=0

Now after we found all three unknown variables, we put them in our original, general quadratic equation.

Numerator=0.5n²+0.5n

Numerator=image09.png

To check, if our expression is correct, we check put in a row number and check if the associated numerator is correct:

Validation test 1:

n=3    

Numerator=(3²+3)/2

  • Numerator=6

Validation test 2:

n=6

Numerator=(6²+6)/2

  • Numerator=21

Validation test 3:

n=7

Numerator=(7²+7)/2

  • Numerator=28

Step 4:

The next step is , to find a general expression for the denominator as a function of row number n and element number r.

image01.png

First of all I’m going to show how the denominator for r=1 (the red circled numbers).

Figure 6 is showing the relation between the denominator and n

To find the general expression for the denominator, we have to use the general, quadratic approach again:

Denominator=an²+bn+c

We need three equations to find all three variables. Therefore we put in the numbers from row 3, 4 and 5 and their associated denominators. Remember that we for now look at the red circled numbers

Quadratic equation for r=1

I:4=9a+3b+c

II:7=16a+4b+c

III:11=25a+5b+c

...read more.

Conclusion

So in line n, r=[0;n]

Step 7:

I will shortly explain, how I arrived at my general statement.

At first, I had a look at the given set of number, where I found patterns between the numerators and the row numbers and also for the denominator and the row numbers.

With that information I came up with the approach using the general quadratic equation. For the numerator it wasn’t too difficult because I only had to solve 3 different equations in order to get the three unknown variables for Numerator (n) =an²+bn+c.

For the denominator, it was a bit trickier. The first step was similar. Using the denominators values to solve Denominator=an²+bn+c. After solving those three equations, I noticed that the first part of the equation looked similar to the general numerator expression. So I tried to make it the same by adding +0,5n-0,5n in order to get 0.5n²+0,5n in each equation. After that was done, I saw that the last part of the equation was always r²-rn.

After finding the formula for the numerator and denominator, finding the general statement for En(r) wasn’t much of a big deal. All I had to do was dividing the numerator by the denominator.

...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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