SL Math IA

Emmanuel Castaño

Ms. Bessette; Period 7

Due: March 30th, 2011

IB SL1 Math Internal Assessment – Lacsap’s Fractions

This assignment required students to find patterns within a triangle of elements which

had noticeable similarities to Pascal’s triangle; a triangle of number’s commonly used for

binomial expansions. An example of Pascal’s triangle can be seen in Figure 1 below:

1 1

1 21

1 331

1 46 4 1

1 510 10 5 1

The variable r is the number of the element in a row, or the number of the column the

element is in. The first element of each row is r=0, and thus r is the (r+1)th of each row. The

equation to find a particular element (E), when n is the row number, and C is a combination from

probability, is:

Er (n )=.n Cr

Row 1

Row 2

Row 3

Row 4

Row 5

Figure 1: These are the first five rows of Pascal’s triangle.

The triangle researched for this internal assessment had the same format as Pascal’s

triangle; however it included fractions and more complex patterns between elements. The first

five rows of the triangle were given, and the students were required to find the elements of the

next two rows. The five rows given were:

1 1

132

1

164

64

1

1107

106

107

1

11511

159

159

1511

1

From this example it can be observed that the numerator of each row remains the same

throughout the whole row, supposing that the first and last term in a row is a fraction. In which

case the first and last terms in the second row two would be 33

, 66

in the third row, and the other

rows would follow this pattern successively. A pattern that can be used to find the numerator of the sixth

and seventh row is shown below:

1 1

1 31

1 66 1

1 1010 10 1

1 1515 15 15 1

Figure 2: These are the first five rows of Lacsap’s Fractions.

Figure 3: This is a simple pattern in numerators of the first five rows of Lacsap’s Fractions.

+3+4

+5

Row 1

Row 2

Row 3

Row 4

Row 5

This pattern shows that the numerator values are growing exponentially in comparison to

the row numbers of the numerators.

This simple pattern can be used to predict that the next to numerators are 21, and 28

respectively. However, to find an equation that could calculate the numerator of the numerator in

the nt h row, a pattern had to be noticed between the row number and the numerator of the row,

instead of the between the numerator from previous rows and the numerator of row n.

To find an equation that could predict the numerator of any row, I set up three systems of

equations and used the method, substitution, to solve for the coefficients.

Row Number (n)

(x )

Numerator

( y )

1 1

2 3

3 6

4 10

5 15

6 21

7 28

As has been previously stated, the relationship between n and the numerator is exponential, and

therefore, the equation for these two sets of values must be a quadratic function. Therefore:

y=A x2+Bx+C

When the values for x=1 in Table 1

are entered:

1=A (1¿¿2)+B(1)+C ¿

Or:

C=−A−B+1


are entered:

Table 1: is a table showing numerators of each row of Lacsap’s Fractions.

3=A (2¿¿2)+B (2)+C ¿

Or:

C=−4 A−2 B+3

By using substitution:

−A−B+1=−4 A−2 B+3

This can be simplified to:

3 A=−B+2

Or:

B=2−3 A

If this is plugged back into𝐶=−𝐴−𝐵+1, then:

C=−A−(2−3 A )+1

Or:

C=2 A−1

If then the values for x=3 are

entered as a quadratic:

6=A (3¿¿2)+B(3)+C ¿

Or:

9 A=−3 B−C+6

If this B and C are then substituted

into 9 A=−3 B−C+6, then:

9 A=−3 (2−3 A )−(2 A−1)+6


9 A=7 A+1

Or:

A=12

From there, it can be plugged into𝐵=2−3𝐴, and C=2 A−1 which returns,

B=12

, and C=0 respectively.

The solution to this system of equations return the values A=12

, B=12

, and C=0. This results

in the quadratic equation, y=12

x2+ 12

x, or y=x (x+1)

2. When x is replaced by the n for the row

number, and y is replaced by the numerator:

Numerator (n)=n(n+1)

2

This equation states that the numerator of any row n will be the row number multiplied

by one more than itself, divided by two.

The table below shows how the numerator of a row can be found with by using

Numerator (n)=n(n+1)

2.

Row Number (n)

(x )

Equation

.y=x (x+1)

2

Numerator

( y )

1.1(1+1)

21

2.2(2+1)

23

3.3(3+1)

26

4.4 (4+1)

210

5.5(5+1)

215

6.6(6+1)

221

7.7(7+1)

228

Table 2: is a table showing how the numerator of each row can be reached from the row number.

The denominator in this triangle is more complicated to predict since it changes between

rows, and in each position within each row. To complete the denominators for the sixth and

seventh row I used the simple pattern shown in Figure 5. By following the pattern shown below,

the denominator can be deduced to be 16, 13, 12, 13, and 16 respectively for r equals 1, 2, 3, 4,

and 5 in the sixth row, and 22, 18, 16, 16, 18, and 22 for r equals 1, 2, 3, 4, 5, and 6 in the

seventh row.

1 1

1 21

1 4 41

1 76 71

1 119 9 111

By noticing the pattern shown in Figure 5 and by using Numerator (n)=n(n+1)

2 for the

numerator, I found the sixth and seventh row of the triangle:

1 1

132

1

164

64

1

1107

106

107

1

11511

159

159

1511

1

12116

2113

2112

2113

2116

1

+2+3

+4

Row 2

Row 3

Row 4

Row 5

Row 1

Figure 5: This is a simple pattern in the denominators of the first five rows of Lacsap’s Fractions.

+2+3 +2

12822

2818

2816

2816

2818

2822

1

Even though the pattern used to find the sixth and seventh row of the triangle is very

simple, to find a general statement for the denominator of the rth term of the nth row, a pattern has

to be found within each row in terms of r. Even though there are many patterns that can be

noticed such as, the middle terms of odd rows can be found by adding one to the row number,

squaring it, and dividing it by two, (n+1)2

2, a useful pattern must work for all rows and all

elements.

I noticed that the numerators are always greater than the denominators which led me to

experimenting with subtracting from the numerator. However, since the denominator decreases

until the middle term, and continues to increase symmetrically as it reaches the end of a row, the

number subtracted from the denominator had to have r2 or |r| to have the subtracted amount be

the same on either side of the middle term in a row.

Then I realized that the pattern for the denominator was quadratic. It started high,

lowered to a vertex, and then increased to its starting point. I also noticed that if the first and last

term of each row follow the pattern that the numerator is the same throughout the whole row,

then to equal, ‘1,’ the first and last term must be the numerator over itself. Therefore, when r=0

the denominator is the same as the numerator, and therefore, y-intercept of the quadratic equation

must be the numerator.

To find an equation that could predict the denominator of any element of any row, I set

up three systems of equations and used the method, substitution, to solve for the coefficients.

Figure 6: These are the first seven rows of Lacsap’s Fractions.

Element Denominator

(r )

Row (n)

( y )

(x ) n=1 n=2 n=3 n=4 n=5 n=6 n=7

0 1 3 6 10 15 21 281 1 2 4 7 11 16 222 3 4 6 9 13 183 6 7 9 12 164 10 11 13 165 15 16 186 21 227 28

As has been previously stated, the relationship between n and the numerator is a

quadratic relationship, and therefore, the equation for these two sets of values must be a

quadratic function. Therefore:

y=A x2+Bx+C

When the values for x=1 for the 3rd

row in Table 3 are entered:

4=A (1¿¿2)+B(1)+C ¿

Or:

C=−A−B+4


are entered:

4=A (2¿¿2)+B(2)+C ¿

Or:

C=−4 A−2 B+4


−A−B+4=−4 A−2 B+4


3 A=−B

Or:

B=−3 A

If this is plugged back into𝐶=−𝐴−𝐵+4 , then:

C=−A−(−3 A)+4

Or:

C=2 A+4

Table 3: is a table showing how the relationship between the number of the element and the denominators in the first seven rows of the triangle.

If then the values for x=3 in Table 3

are entered as a quadratic:

6=A (3¿¿2)+B(3)+C ¿

Or:

9 A=−3 B−C+6



9 A=−3 (−3 A )−(2 A+4)+6


9 A=7 A+2

Or:

A=1

From there, it can be plugged into

B=−3 A, and C=2 A+4 which returns,

B=−3 and C=6 respectively.

The solution to this system of equations return the values A=1, B=−3, and C=6. This

results in the quadratic equation, y=1 x2−3 x+6, or y=x2−3 x+6. When x is replaced by ther

for the number of the element in the row, and y is replaced by the value of the rth element:

Denominator [ Er (n )]=r2−3r+6As predicted, C was equal to the numerator of the nth row

. However, with only one row being tested, a pattern has not being found for the coefficients A

and B. The same method will be used to solve a system of equations for the fourth row of the

triangle.

y=A x2+Bx+C

When the values for x=1 for the 4th

row of the triangle are entered:

7=A (1¿¿2)+B (1)+C ¿

Or:

C=−A−B+7

When the values for x=2 in the

triangle are entered:

6=A (2¿¿2)+B(2)+C ¿

Or:

C=−4 A−2 B+6


−A−B+7=−4 A−2 B+6


3 A=−B−1

Or:

B=−3 A−1

If this is plugged back into𝐶=−𝐴−𝐵+7, then:

C=−A−(−3 A−1)+7

Or:

C=2 A+8

If the values for x=3 in the triangle

are entered as a quadratic:

7=A (3¿¿2)+B(3)+C ¿

Or:

9 A=−3 B−C+7



9 A=−3 (−3 A−1 )−(2 A+8)+7


9 A=7 A+2

Or:

A=1

From there, it can be plugged into

B=−3 A−1, and C=2 A+8 which returns,

B=−4 and C=10 respectively.

The solution to this system of equations return the values A=1, B=−4, and C=10. This

results in the quadratic equation, y=1 x2−4 x+10, or y=x2−4 x+10. When x is replaced by the

r for the number of the element in the row, and y is replaced by the value of the rth element:

Denominator [ Er (n )]=r2−4 r+10 In the 3rd row, the equation to find the values in the

denominator was, r2−3 r+6, and in the 4th row, the equation was r2−4 r+10. In both of these

equations, the A value was, ‘1,’ the B value was equal to the negative row number (−n), and the

C value was equal to the numerator of that row. By following this pattern, the equation for the

denominator of the rth element of the nth row would be:

Denominator=r2−nr+r 0

Where r0 is the numerator, or n(n+1)

2.

Element Denominator

(r )

Row (n)

( y )

(x ) n=1 n=2 n=3 n=4 n=5 n=6 n=7

0 1 3 6 10 15 21 281 1 2 4 7 11 16 222 3 4 6 9 13 183 6 7 9 12 164 10 11 13 165 15 16 186 21 227 28

Table 4 shows the values that will be graphed below. If the equation

Denominator=r2−nr+r 0 is accurate, the graphs, r2−0 r+r0, r2−r+r0,r

2−2r+r 0,r2−3 r+r0,

r2−4 r+r 0, r2−5 r+r0,r

2−6 r+r0, andr2−7 r+r0, should pass through the points of the

corresponding row.

Table 4: is a table showing how the relationship between the number of the element and the denominators in the first seven rows of the triangle.

Figure 7: shows the relationship between the number of the element and the denominator of the first seven rows.

The quadratic equations shown in Figure 1 show that for each one of the first seven rows:

y=1 x2+ yx+x0

When y is replaced with n and x is replaced with r:

Denominator=r2+nr+r0

r0 is also the numerator of the row, or n(n+1)

2:

Denominator=r2+nr+n(n+1)

2

r can be factored out:

Denominator=r (n+1)+n(n+1)

2

Therefore, if r (n+1) is added to the numerator, of any row n, the denominator of term r

can be found. If the equation for the numerator, n(n+1)

2, and the equation for the denominator,

r (n+1)+n(n+1)

2, are combined, it would yield an equation to find any element on any row of

the triangle.

Er (n)=n(n+1)/2

[n (n+1 ) /2 ]+r (r−n)

By using this equation, any term from the triangle below can be found. In the case that

r=0 or r=n, the part of the general equation, r (r−n) will equal to 0 and therefore, the equation

would simplify to, Er (n )=n(n+1)/2n (n+1 )/2

. Therefore the only limitation to this equation is that the

first and last term of each row will equal the numerator over itself. Even though this simplifies to

‘1,’ some of the terms in the triangle, such as E1(3) which equals 64

, are not simplified. With this

in mind, before entering a value into the general equation above, if r=0 or r=n, then, Er (n )=0.

The scope of this equation is for whenever r ≥ 0, and n ≥ 1, when r and n can only be

integers. This is because the triangle’s first element is element 0, and its first row is row 1.

The following values were acquired by using the formula Er (n)=n(n+1)/2

[n (n+1 ) /2 ]+r (r−n).

Row Number (n) Element Number(r ) Equation Element

1 0 1(1+1)/2[1 (1+1 )/2 ]+0 (0−1)

.11

5 2 5(5+1)/2[5 (5+1 ) /2 ]+2(2−5)

.159

7 5 7(7+1) /2[7 (7+1 ) /2 ]+5(5−7)

.2818

8 8 8 (8+1)/2[8 (8+1 )/2 ]+8(8−8)

.3636

11

11

33

32

33

66

64

64

66

1010

107

106

107

1010

1515

1511

159

159

1511

1515

2121

2116

2113

2112

2113

2116

2121

2828

2822

2818

2816

2816

2818

2822

2829

Table 5: is a table showing how the values in “Figure 8”were found.

3636

3629

3624

3621

3620

3621

3624

3629

3636

4545

4537

4531

4527

4525

4525

4527

4531

4537

4545

5555

5546

5539

5534

5531

5530

5531

5534

5539

5546

5555

Figure 8: These are the first ten rows of Lacsap’s Fractions as would be found with

Er (n)=n(n+1)/2

[n (n+1 ) /2 ]+r (r−n).

“I, the undersigned, hereby declare that the following assignment is all my own work and that I

worked independently on it.”

“In this assignment, I used LoggerPro 3.8.2 to draw my graphs.”

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SL Math IA