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Ib Math Sl Type Ii Ia
Lacsap’s Fractions

IB Math SL
Internal Assessment Paper 1

Lacsap’s Fractions Lacsap is Pascal spelled backward. Therefore, Pascal’s Triangle can be used practically especially with this diagram.

(Diagram 1) This diagram is of Pascal’s Triangle and shows the relationship of the row number, n, and the diagonal columns, r. This is evident in Lacsap’s Fractions as well, and can be used to help understand some of the following questions.
Solutions
Describe how to find the numerator of the sixth row. There are multiple methods for finding the numerator of each consecutive row; one way is with the use of a formula, and another by using a diagonal method of counting illustrated by a diagram. The following image can be used to demonstrate both techniques to finding the numerator:

(Diagram 2) This formula uses “n” as the row number and the outcome is the numerator of the requested sixth row. n2 + n
2
As indicated, inputting the number 6, as the requested sixth row, for n gives the solution of 21.
X = n2 + n 2
X = (6)2 + (6) 2
X = 36 + 6 2
X = 42 2
X = 21 Therefore, as shown, the numerator of the sixth row is 21, and this can be checked for validity by entering each number, 1 through 5, into the formula and making sure that the answer corresponds with the numerator in the above diagram.
Where n = 5: Where n = 4:
X = n2 + n X = n2 + n 2 2
X = (5)2 + (5) X = (4)2 + (4) 2 2
X = 25 + 5 X = 16 + 4 2 2
X = 30 X = 20 2 2
X = 15 X = 10
Where n = 3: Where n = 2:
X = n2 + n X = n2 + n 2 2
X = (3)2 + (3) X = (2)2 + (2) 2 2
X = 9 + 3 X = 4 + 2 2 2
X = 12 X = 6 2 2
X = 6 X = 3 As shown, each value of X indicates the corresponding value of the numerator of each row. Another way of finding the

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