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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. 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. 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

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= 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. 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

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.

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

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