1 32 1

1 64 64 1

1 107 106 107 1

1 1511 159 159 1511 1

The aim of this task is to find the general statement for En(r). Let En(r) be the element in the nth row, starting with r = 0.

First to find the numerator of the sixth row, the pattern for the numerator for the first five rows is observed. Since the numerator is the same in each row (not counting the first and the last number in each row), I can observe the numerator in the middle of each row. The numerators from row 1 to row 5 are 1,3,6,10,15 Table 1: A table showing the relationship between the row number and the numerator. The table also shows the relationship between the numerators in each row. Row| Numerator| 1st differences | 2nd differences|

1| 1| 2| 1|

2| 3| | |

| | 3| |

3| 6| | 1|

| | 4| |

4| 10| | 1|

| | 5| |

5| 15| | |

The difference between the numerator in row 1 and row 2 is 2, row 2 and row 3 is 3, row 3 and 4 is 4 and row 4 and 5 is 5. The second difference for each row number is 1; this shows that the equation for the numerator is a geometric sequence. So I try to find the equation of the sequence by using the quadratic formula, y = ax2 + bx + c, where y = the numerator and x = the row number.

6 = a(3)2 + b(3) +0

6 = 9a +3b

6 = 9a + 3(-2a + 1.5)

6 = 9a – 6a + 4.5

1.5 = 3a

a = 0.5

b = -2(0.5) + 1.5

b = -1 + 1.5

b = 0.5

3 = a(2)2 + b(2) +0

3 = 4a +2b

b = 3-4a2

b = -2a + 1.5

First I plugged in 2 for x and 3 for y (the second row) to find the quadratic equation for this sequence. Then I plugged in 3 for x and 6 for y (the third row) because there are 2 unknown variables (“a” and “b”), so I need to 2 equations. Then I substituted “b” from the first equation (the second row)to the second equation (the third row), and then I solve for “a”. I found out that a = 0.5 and b = 0.5. So the equation for the numerator is numerator = 0.5n2 + 0.5n, where n = the row number. By using this equation, the numerator of the sixth row is 0.5(6)2 + 0.5 (6) 0.5(36) + 3 18 + 3 = 21

Row number = 1

0.5(1)2 + 0.5(1) 0.5 + 0.5 = 1

Numerator = 1

Row number = 2

0.5(2)2 + 0.5(2) 0.5(4) + 1 2 + 1 = 3

Numerator = 3

Row number = 3

0.5(3)2 + 0.5(3) 0.5(9) + 1.5 4.5 + 1.5 = 6

Numerator = 6

Row number = 4

0.5(4)2 + 0.5(4) 0.5(16) + 2 8 + 2 = 10

Numerator = 10

Row number = 5

0.5(5)2 + 0.5(5) 0.5(25) + 2.5 12.5 + 2.5 = 15

Numerator = 15

Row number = 6

0.5(6)2 + 0.5(6) 0.5(36) + 3 18 + 3 = 21

Numerator = 21

Figure 1: A figure showing the numerator from row 1 to row 6 by using the equation 0.5n2 + 0.5n to find the numerator. This proves that the equation to find the numerator is accurate.

Then I graph the numerator and the row number on a scatter plot to show the relationship between the numerator and the row number.

Figure 2: A graph showing the relationship between the numerator and the row number. The row number is on the x-axis and the numerator is on the y-axis. This graph has the row number from 1 to 6. According the figure 2 (the graph above), it shows that the numerator increases as the row number increases. The numerator increase at different rate depends on the row number. The trend line is not a linear line, so this shows that the equation for the numerator is not a linear. The trend line for this graph is y = 0.5x2 + 0.5x, which shows that the numerator increases in a quadratic. The R² = 1 shows that the trend line passes through all the points, which illustrates that the equation for the numerator is correct. To find the denominator, I examine the denominator in each row. The denominator is different starting from row 4. The left side of the sequence is symmetry with the right side of the sequence. By looking at each element, I found out that in the first...