Lacsap's Fractions

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Math SL I.A:
Lacsap's Fractions


In my internal assessment, type 1, I was given Lacsap's Fractions task. To do my calculations I used a TI-84 graphic calculator. To type the I.A I used Apple's Pages, Microsoft Excel 2011 and Microsoft Word 2011.

Lacsap's Fractions

To find the numerator of the sixth row I looked at the difference between each of the numerators. 1 1
1 32 1
1 64 64 1
1 107 106 107 1
1 1511 159 159 1511 1

Row| Numerator| Difference between numerators n-nn-1|
1| 1| 1|
2| 3| 2|
3| 6| 3|
4| 10| 4|
5| 15| 5|
6| 21| 6|
7| 28| 7|
From this table one can notice that the difference between each numerator on each row is always d+1, where d represents the difference between the two previous rows. The equation representing this is: un=un-1+(d+1), where un represents the numerator you are looking for, n is the row number and un-1 is the numerator from the previous row. Therefore to find the numerator of the 6th row I did:


After that, I plotted a graph doing numerator vs. row:

The relationship between the numerator and the row is best described by the equation of the line:


The equation is quadratic and can be used to determine any numerator at any row. In the equation, x stands for the row and y represents the numerator. To prove the equation right, I chose three random points:

A (2,3), B (3,6), C (4,10) and plugged it into the equation y=ax2+bx+c



3610=4219311641abc 36100.5-10.5-3.56-2.56-83=abc


Therefore the line equation for the numerator is 12n2+12n=u

The numerator (u) happens to always be bigger than the denominator (v) no matter what row (n) it is present on, therefore one can conclude that the n, v and r are all...
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