Can anyone help me with lascap's fractions?

I did a Google search on Lascap’s fractions and could not find a definition. It seems that this is some kind of proprietary teaching tool. As the topics you chose indicate, the name Lascap seems to be a riff on Pascal. Could you give us the problem you are working on? Maybe we could help you with the answer without knowing what Lascap fractions are.

Lacsap is Pascal backwards. Pascal’s fraction triangle (just an image, not a description) is a way to arrange fractions. Starting with1/1 at the top, the numerator is increased by one each step to downward right, and the denominator increases by one each step to the downward left. Each fraction on the inside has the numerator of the fraction to its upper-right, and the denominator of the fraction to its upper-left. Every rational number appears on this triangle somewhere.

Some additional info, there is a quick and easy way to find the xth number of row y.
x / (y – x + 1)

EDIT: Since Lacsap is probably spelt backwards for a reason, they may want the triangle to be upside-down, or maybe reverse each fraction.

Ok this is how it goes

Aim: In this task you will consider a set of numbers that are presented in a symmetrical pattern.

Basically I’ve been assigned a math project where we with “lascap’s fractions”, which is obviously pascal’s triangle backwards. The diagram below is given.

1 1
1 3/2 1
1 6/4 6/4 1
1 10/7 10/6 10/7 1
1 15/11 15/9 15/9 15/11 1

Basically I’ve been able to find the patterns, like the numerator goes from +2+3+4+5 and so on. A general statement to represent the relationship between the row number n, and the numerator in each row is = .5n(n+1)

The problem that I’m having is finding a general statement that follows these parameters:

Let E n® be the (r+1) element in the nth row, starting with r=0
Example: E 5 (2) = 15/9

@arjen22 Try subtracting the denominator from the numerator in each row.
.....0 0
....0 1 0
...0 2 2 0
..0 3 4 3 0
.0 4 6 6 4 0 etc

To look at this better, I’ll format it differently

0 0
0 1 0
0 2 2 0
0 3 4 3 0
0 4 6 6 4 0 etc

Look at the columns of numbers. The first one is multiples of 0, second multiples of 1, third multiples of 2, etc.

To find the numbers on each row of this triangle, the formula is (C – 1)(R + 1 – C), where R is the row and C is the column.

So, the general formula for the original triangle would be
.5 R (R + 1) / (.5 R (R + 1) – (C – 1) (R + 1 – C))

Which can be shortened to-
R (R + 1) / (2 (C – 1)^2 + R (3 – 2 C + R))

Please do not post further answers here, as this is an IB Assessment tool, and the students are supposed to be showing everything they have learned over the last year through a personal investigation. It is dishonest and unethical for them to seek this help. Thank you!

But what’re you on about, ebywater?? “Students are supposed to be showing everything they have learned over the last year through a personal investigation”?? First of all, we’re not showing anything we’ve learned over the first year of IB. Secondly, this is the pointless task I’ve ever been given in my life!! I mean, you’re wrong by saying that we’re sowing what we’ve learned in Maths. And finally, I find worthless to be doing a task in which we’re not applying the knowledge we’ve learned in IB Maths. And furthermore, we’ve barely been told of what we’re supposed to do with this damned Internal Assessment, and that’s why people are confused and need look for an answer wherever. If you don’t like it, then do not use this website :) Simple is that ;)