# Lacsap’s Fractions Essay Sample

9 September 2017

This diagram is of Pascal’s Triangle and shows the relationship of the row figure. n. and the diagonal columns. r. This is apparent in Lacsap’s Fractions as good. and can be used to assist understand some of the undermentioned inquiries. Solutions

Describe how to happen the numerator of the 6th row.
There are multiple methods for happening the numerator of each back-to-back row ; one manner is with the usage of a expression. and another by utilizing a diagonal method of numbering illustrated by a diagram.
The undermentioned image can be used to show both techniques to happening the numerator:

( Diagram 2 )
This expression uses “n” as the row figure and the result is the numerator of the requested 6th row. n2 + N
2
As indicated. inputting the figure 6. as the requested 6th row. for n gives the solution of 21. X = n2 + N
2
Ten = ( 6 ) 2 + ( 6 )
2
X = 36 + 6
2
X = 42
2
X = 21
Therefore. as shown. the numerator of the 6th row is 21. and this can be checked for cogency by come ining each figure. 1 through 5. into the expression and doing certain that the reply corresponds with the numerator in the above diagram. Where n = 5: Where n = 4:

Ten = n2 + nX = n2 + N
22
Ten = ( 5 ) 2 + ( 5 ) Ten = ( 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:
Ten = n2 + nX = n2 + N
22
Ten = ( 3 ) 2 + ( 3 ) Ten = ( 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 manner of happening the numerator is by sing Diagram 2 and detecting that the ruddy circles environing each numerator. Each subsequent numerator is the add-on of the predating numerator and the current row. For illustration. in row figure 3. the numerator is 6. and the following row figure is 4. therefore. the numerator of row 4 would be the value of 6 + 4. which is so a amount of 10. and can be verified as the numerator of row 4. This method is merely valid with the usage of the first column. as indicated by the missive “r” in Diagram 1. One other restriction of this expression is that it would non work for n = 0. besides known as ‘row zero’ because the “divide by nothing rule” would come into consequence.

Plot the relation between the row figure. 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 figure. n ) . Therefore. the “x” value on the graph ( which in this instance is the numerator of each row ) becomes exponentially larger. doing the gradient. or incline. of the curve to go smaller. until it finally would look to make a incline of nothing. However. that could non be the instance as the incline is greater than zero ( m & gt ; 0 ) . and could non of all time attack nothing. or go a negative incline. because both x and Ys are increasing in the positive way in this state of affairs.

Find the 6th and 7th rows. Describe any forms you used.
The followers is the representation of the first seven rows of Lacsap’s Triangle. excepting the ‘row zero’ as indicated by Diagram 1 of Pascal’s Triangle.

( Diagram 3 )

Forms
Finding the numerators of the 6th and 7th row was the first measure in making the diagram. I used the expression mentioned earlier ( n2 + 2 )
2 in order to find the values of the numerators. The following measure was following similar forms that was used for the numerator. but change it to integrate the denominator part. This was done by adding the denominator in the slant column 1 ( visualized by boxed subdivision of Diagram 3 ) to the row figure. n. numbering the amount of the denominator in the undermentioned row. of the same column. For illustration. in order to acquire the first denominator of the 6th row. I added 11 ( the denominator of the old row ) to the row figure of the 5th row. 5. to acquire the reply of 16. The same was applied to accomplish the first denominator of row 7. As apparent. by analyzing 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 regulations as above mentioned and come to a decision of the denominators of the remainder of the 6th and 7th rows. One other note worth adverting is that the centre triangles in Lacsap’s Fractions are relative fractions. for illustration. 10/6 is the same denary equivalent of 15/9. all three of which form a central-located trigon.

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 figure. r = component satisfies the given illustration and the remainder of Lacsap’s Fractions for the given row sum. This is besides in conformity with the information I presented earlier in the survey. 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 make this expression which serves for a in-between land from which to work away of.

Finding extra rows constitutes stop uping ‘n’ and ‘r’ into the above expression: 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 – [ 7 ]
E8 ( 1 ) = 36
29
Conclusively the informations acquired from this expression. lucifers with the informations obtained through my methods. Therefore. this equation can be used to work out the remainder of the 8th row. and of row nine. and so on. The 8th row:

136/2936/2436/2136/20 36/ 21 36/24 36/29 1 The 9th row:
1 45/3745/3145/2745/2545/2545/ 27 45/31 45/37 1
One bound I noticed when making this. was in row eight. I was convinced or believed that the mirror contemplation theory would work all the manner through Lacsap’s Fractions as it does in Pascal’s Triangle. However. I was incorrect and was later wrong in the analysis of a part of my informations. After a reevaluation of my informations. I discovered a mathematical mistake in my computations and was able to rectify the mistake. which provided a spot of confusion in my overall informations.

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