# Type I Internal Assessment (Lascap's Triangle)

by harshalpatil (student)

## Introduction

In this coursework, we have to look for a pattern in the numbers given to us in a triangular format as seen in the image below.

The image is a Lacsap’s Fraction. Lacsap’s fractions look a lot like Pascal’s Triangle and the work “lacsap” is “pascal” backwards which points further to their relationship.

## Objectives

Our objectives are as follows

1. Find the numerator in the sixth row.
2. Find and plot the relation between the row number and the numerator in each row, write a general formula
3. Find the 6th and 7th rows

Using the patterns in task 3, find the general statement for

. Where

1.  is the (r + 1) th element in the nth row, starting with r = 0.
2. Find additional rows and test if the general statement agrees and is correct
3. Discuss the limitations of the general statement

## Finding the Numerator of the sixth row

By looking at the Lacsap’s fraction and comparing it to the Pascal’s triangle, I observed that the numerators of the Lacsap’s fraction are the same as the third element number of the Pascal’s Triangle as seen in the image below. By this theory, the sixth numerator should be 21.

If we try to find a relationship between the numerators, we see that the difference between each subsequent numerator is 1 in each row starting with 2.

During my research of Pascal’s triangle, I also came across nCr equation, where ‘n’ is the row number and ‘r’ is the element number

For example,

3C2 = (3!) / [( 2!) x (3 - 2)!] = 3

This can be used to find any number on the Pascal’s Triangle considering that the 1 on top of the triangle is represented by n = 0 and r = 0 and the 1 must not be counted as a number when looking for the ‘r’ value.

Therefore this can be used to find the numerators in the Lacsap’s Triangle.

## Finding and plotting the relation between the row number and the numerator in each row and finding a general formula

I have made a table ...