Post on 22-Apr-2015
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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
When the values for x=2 in Table 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
This can be simplified to:
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
When the values for x=2 in Table 3
are entered:
4=A (2¿¿2)+B(2)+C ¿
Or:
C=−4 A−2 B+4
By using substitution:
−A−B+4=−4 A−2 B+4
This can be simplified to:
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
If this B and C are then substituted
into 9 A=−3 B−C+6, then:
9 A=−3 (−3 A )−(2 A+4)+6
This can be simplified to:
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
By using substitution:
−A−B+7=−4 A−2 B+6
This can be simplified to:
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
If this B and C are then substituted
into 9 A=−3 B−C+7, then:
9 A=−3 (−3 A−1 )−(2 A+8)+7
This can be simplified to:
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.”