Introduction
In this activity, set of numbers which are presented in symmetrical pattern are considered in each case, a general formula is found for the elements in each and every row. Below there are a set of numbers which we will consider in this activity.




1 

1 



Row1 



1 

3/2 

1 


Row2 


1 

6/4 

6/4 

1 

Row3 

1 

10/7 

10/6 

610/7 

1 
Row 4 
1 

15/11 

15/9 

15/9 

15/11 

1Row5 
This is figure 1 showing Lacsap’s fraction
Plotting the graph using technology
From the data, a graph to identify the relationship between row number (r) and the numerator in each and every row was drawn as below.
Figure 2: This figure shows the relationship between row number and the numerator value of Lacsap’s fraction.
When the graph is analyzed carefully, it is clear that the numerator value in each and every row is consistent in the case where all the number 1’s are discarded. To find the numerator of the sixth row, the sequence of the numerator in each row can be observed. Discarding all the number 1’s makes the sequence to begin at row 2 and not as in the initial sequence where it started at row 1. The numerator sequence includes 3, 6, 10, and 15. From this sequence, its is clear that the numerator increase by adding a factor of one i.e. the numerator is increased by 1 as compared to the immediate previous value. When the sequence from 3 to 6 is considered, it can be concluded that the numerator is increased by 3. This process continues as in 6 t0 10 where the numerator is increased by 4.
From this, an equation is derived using the row number which eventually is utilized as a variable that calculates the numerator. In using the row number as the base to calculate the numerator, it can be observed that getting from 2 – row number 2  to 3 – row number 3 – is achieved by multiplying the row number by a value of 1.5. Also, in the next row, getting from row number 3 to row number 6 the row number is multiplied by a value of 2. Likewise, in row 4 to 10, the row number is multiplied by 2.5. From this observation, it can be concluded that getting from one row to the immediate next row, the row number is multiplied by a value which is 0.5 more than the previous row. This can be formulated into an equation, whereby each and every row number – designated by r, is multiplied by row divided by 2 and eventually adding the 0.5. In summary, this can be represented in equation 1 below.
Equation………………………………..I
Finding the numerator of the sixth row
To get the numerator of the sixth row, the equation formulated above is used.
Want an expert to write a paper for you Talk to an operator now…………………………Equation ii
Since it is the sixth row, we replace r with 6
= 21
Finding the sixth and the seventh rows, describing patterns used and getting the general statement for E_{r} (n)
In addition to the numerator of each row, the denominator of each row has to be found in determination of each element in a row. This can be achieved through observing the relationship between each row’s numerator and the corresponding denominator. As In the case above, the 1’s are discarded and thereafter, getting the denominator of each row involves observing the first element of the row as indicated in the diagram below.




1 

1 



Row1 



1 

3/2 

1 


Row2 


1 

6/4 

6/4 

1 

Row3 

1 

10/7 

10/6 

610/7 

1 
Row 4 
1 

15/11 

15/9 

15/9 

15/11 

1Row5 
Figure 3 showing relationship between numerator and denominator
The yellow highlighting represents the observed difference between the numerator and the denominator and it was recorded as 1, 2, 3 and 4 for the respective rows of 2, 3, 4, and 5. As a result, the numerator was used as a base to find the denominator as in the initial equation below.
Numerator – y = denominator ……………………………..Equation iii.
This is where y represents the difference between the denominator and the numerator which is formed by using the row and element number. The fact that difference of the denominator of element 1 increases by 1 every consecutive row helps construct a statement which states that subtracting 1 from the numerator and then multiplying by row number minus 1 gives the he denominator of the element. This can be shown in equation iii below where
…………………………Equation iv
The blue highlighting indicates the second row. In this row the difference was recorded as 2, 4, and 6. In element two, this change was an incremental of 2 in every consecutive row. This from this, the formula written below was constructed.
– n(r1) = denominator………………………………..v
Also, the formula equals to
Numerator – n(r – 1) = denominator.
Combining the general statement of the numerator and the denominator, the sixth and the seventh row can be gotten. This is illustrated in the equation VI below
Using the equation and substituting r to be the row number and starting at 2 and n for the element number, where n starts at 1 up to n=  1 , we can get the sixth row as:
Thus the sixth row is
21/16 21/13 21/12 21/13 21/16
Since at the beginning of the calculations the 1’s were discarded, we can replace them back and have the following values.
1 21/16 21/13 21/12 21/13 21/16 1
Like wise the seventh equation is calculated in the same way and the results gotten are as below
1 28/21 28/18 28/16 28/16 28/18 28/21 1
Testing validity of the general statement
The technique used to get the sixth and the seventh method can be used to calculate other rows i.e. the eight, the ninth and the tenth.
1 36/29 36/24 36/21 36/20 36/21 36/24 36/29 1
1 45/37 45/31 45/27 45/25 45/25 45/27 45/31 45/ 37 1
1 55/46 55/39 55/34 55/31 55/30 55/31 55/34 55/39 55/46 1
Scope and limitations of the general statement
In the process of finding the numerator and the denominator, the 1 was discarded. This implies that it the general formula can not be used to calculate the values of the first row thus it’s a limitation. Moreover, the discarding of 1 decrease the number of each row by two thus the element number does not begin at zero but begins at one.
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