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

*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) from n=1 until n=5, again with red numbers representing n:

1. 1        1
2. 1                1
3. 1                        1
4. 1                                1
5. 1                                        1

If you will notice, row 1 (n=1) has a numerator of 1; row 2 has a numerator of 3; row 3 has a numerator of 6, row 4 has a numerator of 10, and row 5 has a numerator of 15.

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. You will find the number 6.

I will demonstrate this when n=4 and n=5 in Lacsap’s Fractions as well and also when n=6, which was not included in the given symmetrical pattern.

When n=4, 4+1=5. Find n=5 in Pascal’s Triangle in c2 and you will see the number 10.

When n=5, 5+1=6. Find n=6 in Pascal’s Triangle in c2 and you will see the number 15.

When n=6, 6+1=7. Find n=7 in Pascal’s Triangle in c2 and you will see the number 21.

Therefore, the numerator of the sixth row would be 21.

Now I will plot the relations between the row number, n, of the given symmetrical pattern “Lascap’s Fractions” and it’s numerator.

The pattern of the plotted points is clearly not linear.

 A linear line is y = mx + b

To find m we would need the distance between two points, or slope.

The distance between the points when n=1 and n=2:

However, the distance between the points when n=2 and n=3:

If the equation were linear then the slope would remain consistent.

Therefore I will go past 1st differences and look at 2nd differences to determine whether or not this is a quadratic equation.

Since the second differences are the same, I am able to understand that this is a quadratic equation.

Formula for quadratic equation: y = ax2 + bx +c where a≠0.

Now I will plug in x and y values for the equation using this chart:

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        1        2        3        4        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.