Rahat Madarasmi

Math Internal Assessment: Lascap’s Fractions

The aim of the assessment is to find, with the use of technology, a general statement. This general statement will include an equation to find the relationship between the row number (n) and the numerator in each row and an equation to find the relationship between the row number, the term number (r) and the denominator in each term.

The pattern that is present is known as Lascap’s Fractions:

1 1

1 3/2 1

1 6/4 6/4 1

1 10/7 10/6 10/7 1

1 15/11 15/9 15/9 15/11 1

The formula that was given was to be used like this: Let En(r) be the (r+1)th element in the n th row, starting with r=0. An example of this would be E5(2)=15/9

In order to find the numerator for the next row, a pattern had to be found between the given numerators. The numbers had to be looked at from the triangle diagonally. By diagonally, this means looking at the numerators like this:

1 1

1 3/2 1

1 6/4 6/4 1

1 10/7 10/6 10/7 1

1 15/11 15/9 15/9 15/11 1

When it is looked at like this, then you list out the numerators to begin finding a pattern:

1 3 6 10 15

Once the numerators are listed out, you then begin to find a pattern between the numbers. Starting by looking for common differences:

1 3 6 10 15

\/ \/ \/ \/

2 3 4 5 1st difference

It can be seen from differences that the 1st differences are not equal, therefore not presenting a pattern. Due to this, a second difference must be found in the attempt to find a common difference:

1 3 6 10 15

\/ \/ \/ \/

2 3 4 5 1st difference

\/ \/ \/

1 1 1 2nd difference

It can be seen from the differences that the 2nd differences are indeed equal therefore presenting a pattern. It is a rule that if the 1st differences are equal, then the function for the numerator will be linear but if the 2nd differences are equal, then the function will be a quadratic. Now to solve the pattern and reach a quadratic function, let us begin by giving numbers to each row of differences:

1 3 6 10 15 row 3

\/ \/ \/ \/

2 3 4 5 row 2 (1st diff.)

\/ \/ \/

1 1 1 row 1 (2nd diff)

The general statement should be in the form of:

tn = an2 + bn + c

In order to find a, b and c, equations must be made to solve for each variable. There are three rules for these three equations. The equation for row 1 to solve for “a” must be: 2a=1 (the first number in row 1). The equation for row 2 to solve for “b” must be: 3a+b=2 (the first number in row 2). An the equation for row 3 to solve for “c” must be: a+b+c=1 (first number in row 3). The figure below will demonstrate this:

a+b+c1 3 6 10 15 row 3

\/ \/ \/ \/

3a+b 2 3 4 5 row 2 (1st diff.)

\/ \/ \/

2a 1 1 1 row 1 (2nd diff)

The bold numbers represent the first numbers in the rows and are what the functions are equal to. Now to solve for each variable, here is the math:

2a=1 3a+b=2 a+b+c=1

a=½ 3(1/2)+b=2 (1/2)+(1/2)+c=1

b=2-(3/2) 1+c=1

b= ½ c=0

To describe the procedure above: after having solved for “a”, the value was substituted into the equation that was used to solve for “b”. After solving for “b”, that value and the value for “a” was substituted into the equation that was used to solve for “c”. In the end, the values for the variables were: a=1/2; b=1/2; c=0. This results in the equation:

tn = ½n2 + ½n

Now in order to get the final equation for the numerator, this equation had to be simplified. So the final equation is:

tn= ½n(n+1)

To test this equation, since it is known that the numerator in the fifth row is 15, 5 will be substituted for “n” as it is the row number:

tn= ½n(n+1)

tn= ½(5)(5+1)

tn= ½(5)(6)

tn= ½(30)

tn= 15

To find the numerator in the sixth row, the number 6 will be substituted for “n” as it is the row number:

tn= ½n(n+1)

tn= ½(6)(6+1)

tn= ½(6)(7)

tn= ½(42)

tn= 21

The numerator in the sixth row will be 21 according to the equation that was used to solve the problem.

There is an alternative method that can be used. The one that was done just now was the manual method, but there is also a technological method that can be used with the calculator.

When entering the equation that was achieved through the manual method into the calculator, a graph like this will appear that proves that it is a quadratic equation.

As this is a quadratic graph, we can use the calculator to solve the equation by using the “Quadratic Regression (QuadReg)” function. The steps to doing this are as follows:

Begin by pressing the “stat” button on your calculator then go to “edit” and click enter.

Once you have gone into “edit”, you will see L1 and L2. In the column L1, enter in all the row numbers from 1-5. In the column L2, enter in all the numerators from rows 1-5 (1,3,6,10, 15).

Once this is done, exit the screen to the main screen so that the screen is blank again. Now press the “stat” button again and scroll right to “calc”.

Now scroll down to “QuadReg” and hit enter.

Once you hit enter, your screen will look like this:

Once you’ve seen that on your screen, hit enter again and your equation will instantly come up.

As solved manually as well, the equation has come out to be the same on the calculator:

tn = ½n2 + ½n

Which then simplifies to become your final equation for the numerator:

tn= ½n(n+1)

In order to find the denominator, technology can be used in the form of the Logger Pro program and graphing.

In order to attain the equation for the denominator, the data that was given in “Lascap’s triangle” had to be entered into the Logger Pro program and a graph had to be created. This graph then had to be analyzed to get the equation.

The data that was entered in to create the graph was all the denominators in each row versus their term numbers. In this case, as row one was ignored, there were 4 lines on the graph. This was because only rows 2 through 5 were graphed. From this graph, a series of equations emerged that would lead us to the final denominator equation. But first, a pattern between the equations had to be found in the form of: ax2 + bx + c.

Equation Row 2: x2 + (-2x) + 3

Equation Row 3: x2 + (-3x) +6

Equation Row 4: x2 + (-4x) +10

Equation Row 5: x2 + (-15x) +15

It can be seen that the common pattern between these equations are that a is always equal to 1, b is always –n, and c is always the first denominator in that given row. So from these conclusions, an general equation can be made to find the denominator. This equation is:

r2 – nr +r0

Where r0 represents the first denominator of the row. Since r0 is a limitation, in order to find r0 the equation for the numerator must be substituted in. So in this case, the equation would look like this:

r2 – nr +(½n(n+1))

Now in order to check that this equation is indeed valid, term 1 from row 4 will be substituted into the equation to see if the results are the same. The given denominator for this term is 7.

r2 – nr +(½n(n+1))

r2 – 4r +(½(4)(4+1))

r2 – 4r +(2 (5))

(1)2 – 4(1) +10

1-4+10

=7

Now that, both, the equation for the numerator and the denominator have been found, in order create a final equation that can be used to find every single term in every row, both of these equations need to be combined. This equation is the final equation that solves this triangle and can be used to find every term that is desired in this pattern. This is the final equation:

(½n(n+1)) / (r2 – nr +r0)

To prove that this equation works, a given term can be substituted into the equation to see whether the results are the same. Term 3 in row 5 will be substituted in. The given answer is 15/9.

(½n(n+1)) / (r2 – nr +r0)

(½(5)(5+1)) / (32 – (5)(3) + (½n(n+1)))

((2.5)(6)) / (9 – 15 + (½(5)((5+1)))

15 / (9 – 15 +15)

=15/9

It can be seen that this does indeed apply to this pattern of numbers and can be used to solve for any term.

Some of the limitations are that the term number must always be greater than zero and they also must be less than n.

Analysis:

In this internal assessment for math, I used both perception and reason in order to be successful. It was important that I used both coupled together to get successful results. Primarily, in order to recognize a pattern, both reason and perception were used. Perception was used because it was necessary to perceive the fact that there was a pattern among the numbers. Reason was then used to put together the correct numbers that formed the pattern. Based on given knowledge in the directions, I was able to perceive that there was a pattern among the numbers. My reasoning was a product of this perception as I used deductive reasoning to deduce the numbers that would create a pattern. After finding a pattern, perception allowed me to know that there was an equation to come out of this pattern and then reason allowed me to find this pattern. From given information, there was to be an equation out of this pattern and thus, reason was used to turn the pattern into an equation. In order to solve this problem presented in the internal assessment, perception was necessary in order to know what the product was to be, and from this perception, reason was used to explain the procedure as to how to arrive at this perceived conclusion.