Lacsap triangle investigation.

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MATH PORTFOLIO SL

Lacsap’s Fractions

Meher Vaswani

The numbers given in the task sheet bore similarity to Pascal’s triangle. But what gave me certainty was the fact that ‘Lacsap’ is actually ‘Pascal’ spelt backwards.

Named after a famous French mathematician and philosopher, Blaise Pascal, the Pascal triangle involves the use of many important patterns.

For our assessment, we aim to find the general statement of the element of R in the nth row or En(r) where r=0.

In the given task sheet, we have been given the fractions in the form:


Let us first observe the numerators given. Comparing it to Pascal’s triangle, we are able to deduce that there is a similarity in pattern.

 


The numbers found in the diagonal starting at row 3 or the third element (from the left) in each row starting from row 3, will give us the numerators. For instance, the first number is 1, the second is 3, the third is 6, the fourth is 10 and the fifth is 15. This is known as triangular numbers. Triangular numbers are just one type of polygonal numbers.

"Pascal's Triangle." All You Ever Wanted to Know About and More. Web. 27 Apr. 2012. <http://ptri1.tripod.com/>.





Therefore, the numerator for the 6
th row can be assumed to be 21. The 7th row can assumed to be 28 and the 8th row can be assumed to be 36. Let us now unfold the mystery behind this pattern by first finding the common difference.


The 1st difference between the numerators does not give us a consistent number thus; there is no consistent increase. However, through the 2nd difference, we get the common difference of 1. Therefore, we can plug it into a quadratic formula; y= ax2+bx+c where y = numerator and x = row number. 

When we plug in values into this equation, we can obtain a general formula for the numerator. Let us take the 3rd row with the numerator of 6.

Y= ax2+bx+c

6= a (3)2+b(3)+ 0
6= 9a + 3b
6 – 9a = 3b
6 – 9a = b
    3
2 – 3a = b

Using the 4th row with the numerator 10, we substitute the value of ‘b’ into this second equation

10 = a (4)2+(2-3a)(4)+0
10 = 16a + (8 – 12a)
10 =  4a + 8
10 – 8 = 4a
2 = 4a
1 = a
2
0.5 = a

Once we have found the value of ‘a’ and an equation for ‘b’, we can easily replace the value of ‘a’ in the equation of ‘b’.

2 – 3(0.5) = b
2 – 1.5 = b
0.5 = b

Now when we combine the values of ‘a’ and ‘b’, we get the quadratic equation:

0.5n2 + 0.5n written in terms of x and y is y= 0.5x2 + 0.5x, where y is the numerator value and x is the row number.

Let us prove this equation using the 4th row with the numerator value 10
y =0.5(4)
2  + 0.5(4)
y =0.5(16) + 2
y =8 + 2
y =10

Join now!

5th row with the numerator value 15
y =0.5(5)
2  + 0.5(5)
y =0.5(25) + 2.5
y =12.5 + 2.5
y =15

Why don’t we try to use this formula to find the next several rows to see how accurate and reliable it really is?

According to the triangular numbers, row 6 should have a numerator of 21.
y =0.5(6)
2  + 0.5(6)
y =0.5(36) + 3
y =18 + 3
y =21

According to the triangular numbers, row 7 should have a numerator of 28.
y=0.5(7)
2  + 0.5(7)
y=0.5(49) + 3.5
y=24.5 + 3.5
y= 28

According to the triangular numbers, row 8 should have the numerator of 36.
y=0.5(8)
2 + 0.5(8)
y=0.5(64) + 4
y=32 ...

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