Lacsap's Triangle Maths Portfolio
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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
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