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# Sl Math Ia – Lacsap’s Fraction Essay

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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.

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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 |3 | |5 |4 | The difference in the first element (r=1) is always 1 less than the row number (n). So I can say that: Denominator (r1) = Numerator (r1) – (n-1) Or Denominator (r1) = [pic] Row number (n) |[pic] |Denominator | |2 |(2+1)C(2)-(2-1) |= 2 | |3 |(3+1)C(2)-(3-1) |= 4 | |4 |(4+1)C(2)-(4-1) |= 7 | |5 |(5+1)C(2)-(5-1) |= 11 | This statement will only work for the fractions in the first element (r1) of the triangle. The difference between the difference of Num. and Denom. and the value of ‘n’ in the second element is 2. Row number (n) |Difference between numerator and denominator in the 2nd element | |1 |/ | |2 |0 | |3 |2 | |4 |4 | |5 |6 | After some trial and error I found that the connection between ‘n’ and the difference between Num. and Demon. Is 2(n-2) Row number (n) |[pic] |Denominator | |2 |(2+1)C(2)-2(2-2) |= 0 | |3 |(3+1)C(2)-2(3-2) |= 4 | |4 |(4+1)C(2)-2(4-2) |= 6 | |5 |(5+1)C(2)-2(5-2) |= 9 | Now we have a statement for the 1st and 2nd element it is possible to see a pattern between them. [pic] can also be written as [pic] So First element = [pic]

Second element = [pic] From this pattern we can derive that the general formula for the denominator of any element in any row will be: [pic] Sample: When ‘n’ =5 and ‘r’ = 2[pic] 15 – 6 = 9 When ‘n’ =4 and ‘r’ =2 [pic] 10 – 4 = 6 So now we can conclude that the general statement to finding the fraction in lacsap’s triangle is: [pic] Now that we have the general statement we can use it to calculate the fractions in the 6th and 7th row of the triangle. Row 6: 1st element [pic] 2nd element [pic] 3rd element [pic] 4th element [pic] 5th element Row 6: /// [pic] Row 7: [pic] So to conclude, I was able to find that the general statement of lacsap’s fractions is [pic]

I found this formula by relating the fractions to Pascal’s triangle after realizing that Lacsap was Pacal spelt backwards, from then it was just taking logical steps and using a bit of trial and error to get a formula that will work for any position in the triangle. Some of the things that limit this equation is the fact that all the number 1s must be ignored to get the correct ‘r’ value. This means that the second element in each row will become the 1st element (r=1) in order for the general statement to be true. Also the equation is only valid when ‘n’ and ‘r’ are greater than 0. ———————– Row 7: 1st element [pic] 2nd element [pic] 3rd element [pic] 4th element [pic] 5th element [pic] 6th element [pic] [pic]

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