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This diagram is of Pascal’s Triangle and shows the relationship of the row number, n, and the diagonal columns, r. This is evident in Lacsap’s Fractions as well, and can be used to help understand some of the following questions. Solutions

Describe how to find the numerator of the sixth row.
There are multiple methods for finding the numerator of each consecutive row; one way is with the use of a formula, and another by using a diagonal method of counting illustrated by a diagram.
The following image can be used to demonstrate both techniques to finding the numerator:

(Diagram 2)
This formula uses “n” as the row number and the outcome is the numerator of the requested sixth row. n2 + n
2
As indicated, inputting the number 6, as the requested sixth row, for n gives the solution of 21. X = n2 + n
2
X = (6)2 + (6)
2
X = 36 + 6
2
X = 42
2
X = 21
Therefore, as shown, the numerator of the sixth row is 21, and this can be checked for validity by entering each number, 1 through 5, into the formula and making sure that the answer corresponds with the numerator in the above diagram. Where n = 5:Where n = 4:

X = n2 + nX = n2 + n
22
X = (5)2 + (5)X = (4)2 + (4)
2 2
X = 25 + 5X = 16 + 4
2 2
X = 30X = 20
2 2
X = 15X = 10
Where n = 3:Where n = 2:
X = n2 + nX = n2 + n
22
X = (3)2 + (3)X = (2)2 + (2)
2 2
X = 9 + 3X = 4 + 2
22
X = 12X = 6
2 2
X = 6X = 3

As shown, each value of X indicates the corresponding value of the numerator of each row.
Another way of finding the numerator is by viewing Diagram 2 and noticing that the red circles surrounding each numerator. Each subsequent numerator is the addition of the preceding numerator and the current row. For example, in row number 3, the numerator is 6, and the next row number is 4, thus, the numerator of row 4 would be the value of 6 + 4, which is indeed a sum of 10, and can be verified as the numerator of row 4. This method is only valid with the use of the first column, as indicated by the letter “r” in Diagram 1. One other limitation of this formula is that it would not work for n = 0, also known as ‘row zero’ because the “divide by zero rule” would come into effect.

Plot the relation between the row number, n, and the numerator in each row.

(Graph 1)
Analyzing the graph, it can be seen that the distance between each point becomes larger, the larger the “y” value becomes (the y value is indicated by the row number, n). Therefore, the “x” value on the graph (which in this case is the numerator of each row) becomes exponentially larger, causing the gradient, or slope, of the curve to become smaller, until it eventually would seem to reach a slope of zero. However, that could not be the case as the slope is greater than zero (m > 0), and could not ever approach zero, or become a negative slope, because both x and y are increasing in the positive direction in this situation.

Find the sixth and seventh rows. Describe any patterns you used.
The following is the representation of the first seven rows of Lacsap’s Triangle, excluding the ‘row zero’ as indicated by Diagram 1 of Pascal’s Triangle.

(Diagram 3)

Patterns
Finding the numerators of the sixth and seventh row was the first step in creating the diagram. I used the formula mentioned earlier (n2 + 2)
2 in order to determine the values of the numerators. The next step was following similar patterns that was used for the numerator, but change it to incorporate the denominator portion. This was done by adding the denominator in the slant column 1 (visualized by boxed section of Diagram 3) to the row number, n, totaling the sum of the denominator in the following row, of the same column. For example, in order to get the first denominator of the sixth row, I added 11 (the denominator of the previous row) to the row number of the fifth row, 5, to get the answer of 16. The same was applied to achieve the first denominator of row 7. As evident, by studying row four and five of Lacsap’s Fractions, and the same for Pascal’s Triangle, one can see that the denominator of rows six and seven will be mirror images after r = 3. We can follow the same rules as above mentioned and come to a conclusion of the denominators of the rest of the sixth and seventh rows. One other note worth mentioning is that the center triangles in Lacsap’s Fractions are proportional fractions, for example, 10/6 is the same decimal equivalent of 15/9, all three of which form a central-located triangle.

Find the general statement for En(r).
The Formula 0.5n2 + 0.5n
0.5n2 + 0.5n – [r(n-r)]where r ≠ 0, n = row number, r = element satisfies the given example and the rest of Lacsap’s Fractions for the given row amount. This is also in accordance with the data I presented earlier in the study. I used the numerator equation that I explained earlier, but condensed it so it would be easier to understand. The denominator equation was found on the Internet, and I combined the two, to create this formula which serves for a middle ground from which to work off of.

Finding additional rows constitutes plugging ‘n’ and ‘r’ into the above formula: En(r) = 0.5n2 + 0.5n
0.5n2 + 0.5n – [r(n-r)]
E8(1) = 0.5(8)2 + 0.5(8)
0.5(8)2 + 0.5(8) – [1(8-1)]
E8(1) = 32 + 4
32 + 4 – 
E8(1) = 36
29
Conclusively the data acquired from this formula, matches with the data obtained through my methods. Therefore, this equation can be used to solve the rest of the eighth row, and of row nine, and so on. The eighth row:

136/2936/2436/2136/20 36/ 21 36/24 36/29 1 The ninth row:
1 45/3745/3145/2745/2545/2545/ 27 45/31 45/37 1
One limit I noticed when doing this, was in row eight. I was convinced or believed that the mirror reflection theory would work all the way through Lacsap’s Fractions as it does in Pascal’s Triangle. However, I was wrong and was subsequently incorrect in the analysis of a portion of my data. After a reevaluation of my data, I discovered a mathematical error in my calculations and was able to correct the error, which provided a bit of confusion in my overall data.

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