Sl Math Ia - Lacsap's Fraction

Topics: Fraction, Number, Elementary arithmetic Pages: 7 (852 words) Published: November 14, 2012
Lacsap’s Fractions

The aim if this IA is to investigate Lacsap’s Fractions and to come up with a general statement for finding the terms.

When I noticed that Lacsap was Pascal spelt backwards I decided to look for a connection with Pascal’s triangle.

Pascal’s triangle is used to show the numbers of ‘n’ choose ‘r’(nCr). The row number represents the value of ‘and the column number represents the ‘r’ value. Eg. Row 3, colomn 2 = 3C2 = 2.

I noticed that all the numerators of the fractions in Lascap’s fraction (3,6,10,15) are also found in Pascal’s triangle. So I tried to see if I would get the denominator of the fractions by using the row as ‘n’ and the colomn (or element) as ‘r’. This did not work out because Lascap’s triangle does not have a row with only one element like Pascal’s does. To solve this I just added 1 to each row number. This gives me the formula[pic].

|(Row number +1)C2 |Numerator | |(2+1)C2 |= 3 | |(3+1) C2 |= 6 | |(4+1)C2 |=10 | |(5+1)C2 |=15 |

Now that we have found an equation to solve to numerator of the fractions, we now have to try and work out how to solve the denominator. As we already know the numerator, we can presume that there is some sort of relationship between this and the denominator.

I noticed that the difference between the numerator and denominator increased by one with each row I went down (1,2,3,4)

|Row number |Numerator (2nd element) |Denominator | |1 |/ |/ | |2 |3 |2 | |3 |6 |4 | |4 |10 |7 | |5 |15 |11 | |6 | | |

Because the denominators in one row are different from each other (row 3 = 7, 6, 7), it can be presumed that the unknown ‘r’ which stands for the elements/column will be part of the general statement.

Because the first and last element in each row is 1 (or 1/1) and no general statement is needed to find these, they can be ignored and erased from the triangle for now.

|Row number (n) |Difference between numerator and denominator in the 1st element (see | | |above) | |1 |/ | |2 |1 | |3 |2 | |4...
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