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# Math IA type I. Here is Lacsaps Fractions (the symmetrical pattern given) from n=1 until n=5, again with red numbers representing n:

Extracts from this document...

Introduction

Lacsap’s Fractions

A Math Internal Assessment

By Kelsey Kennelly

Lacsap’s Fractions…Ha

Clever, IB. You took the word “Pascal” and spelled it backwards to make “Lacsap”. I thought this was simply a Math Assessment but I guess it’s also a word scramble.

Anyways, with that being said, here is Pascal’s Triangle starting at n=0 until n=9, with red numbers representing n:

1. 1
2. 1        1
3. 1        2        1
4. 1        33        1
5. 1        4        6        4        1
6. 1        5        1010        5        1
7. 1        6        15        20        15        6        1
8. 1        7        21        35        35        21        7        1
9. 1        8        28        46        70        46        28        8        1
10. 1        9        36        84        126        126        84        36        9        1

*Column         0        1        2        3        4        5        6        7        8        9

*Columns rise diagonally from left to right. Column numbers will be represented by “c” followed by a subscript number.

Here is “Lacsap’s” Fractions (the symmetrical pattern given)

Middle

The numbers of these numerators can be found in the above example of Pascal’s Triangle (column 2 or c2; or c7 but we’ll focus on c2). The number 1 can be found when n=2 in the second column (c2) of the triangle, which is the numerator of row 1 in Lacsap’s Fractions. Furthermore, when n=3 in the second column (c2) of Pascal’s Triangle, the number 3 is found, which is the numerator in Lacsap’s Fractions when n=2. This pattern continues, which brings me to the idea that the numerator is found by adding 1 to n in Lacsap’s fractions and finding that number in c2 of Pascal’s Triangle. For example, when n=3 in Lacsap’s Fractions, the numerator is 6. When you add 1 to 3, the result is 4; find n=4 in Pascal’s Triangle and find column 2.

Conclusion

a =

a=   =

I used the quadratic equation to solve for row 2 and row 4. I used substitution to find the missing variables a and b to arrive at a general statement of

y = x2 + x

where y is the numerator and x is the row number (n)

I have already discovered that the numerator when n=6 is 21.

y = (6)2 + (6)

y = (36)+ (6)

y = 18 + 3

y = 21

Now I will test the statement when n=7.

y = (7)2 + (7)

y = (49)+ (7)

y= 24.5 + 3.5

y = 28

This is correct because when n=7, 7+1=8. Find n=8 in Pascal’s Triangle in c2 and you will see the number 28. Now that I have discovered the general statement for the numerator, it is time to work on the denominator.

1        1

1                1

1                        1

1                                1

1                                        1

Column        1234        5

There is a pattern between the differences of numerator and denominator for each row. In row 1, for example,  is displayed and there is a difference of zero for numerator and denominator.  This continues in the symmetrical pattern.

 Row Column Numerator – Denominator 1 1 0 2 1

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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