Sl Math Internal Assessment: Stellar Numbers

Topics: Real number, Polynomial, Addition Pages: 16 (4160 words) Published: February 24, 2013
SL Math Internal Assessment: Stellar Numbers

374603
Mr. T. Persaud
Due Date: March 07, 2011
Part 1: Below is a series of triangle patterned sets of dots. The numbers of dots in each diagram are examples of triangular numbers. Let the variable ‘n’ represent the term number in the sequence. n=1 n=2 n=3 n=4 n=5

1 3 6 10 15 n=6 n=7 n=8

21 28 36 Term Number (n)| Number of Dots (tn)| First Difference| Second Difference| 1| 1| -| -|

2| 3| 2| -|
3| 6| 3| 1|
4| 10| 4| 1|
5| 15| 5| 1|
6| 21| 6| 1|
7| 28| 7| 1|
8| 36| 8| 1|
As we can see from the chart above, there is a growing increase in the differences between each consecutive set of numbers of dots. The difference between the number of dots in sets 2 and 1 is 2, the difference between the number of dots in sets 3 and 2 is 3, the difference between the number of dots in sets 4 and 3 is 4, and so on. The differences between each consecutive pair of triangular numbers increase constantly by 1 each time. This dictates that there is a constant second difference of 1 for this sequence. The term values, or in this case the number of dots, increase neither arithmetically nor geometrically. This piece of information, shows that the general statement will require an exponent on the variable ‘n’ in order to define a relationship between the term number and the term value. The constant second differences indicate that ‘n’ will be raised to the exponent 2. Differences can be seen to represent derivatives. The second difference represents the second derivative, where the exponent on the primary variable has been reduced by 2, having been differentiated twice. In the same manner, the first difference represents the first derivative where the exponent on the primary variable has been reduced by just 1. In order to return to the primary equation, the exponent from the second derivative equation will have to be raised twice, which is why the exponent on ‘n’ will be 2 and the general statement will be in the form of a quadratic.

The general form for a quadratic is: y=ax2+bx+c
This being a general statement for a sequence, and our primary variable being ‘n’, we will use the general form: tn=an2+bn+c where tn is the term value (number of dots) and ‘n’ is the term number.

In order to determine the general statement of the sequence, we can simply solve for the values of the constants ‘a’ ‘b’ and ‘c’. In order to do so, we will need to define 3 equations and use the method of elimination to solve for the general statement. Elimination requires adding or subtracting 2 equations in the aim of eliminating a constant/variable by first preparing them to share the same coefficients with either the same or opposite integers (depending on whether you will be adding or subtracting the equations).

Equation #1 - For term 1: t1=an2+bn+c
1=12a+1b+c
1=a+b+c
Equation #2 - For term 2: t2=an2+bn+c
3=22a+2b+c
3=4a+2b+c
Equation #3 - For term 3: t3=an2+bn+c
6=32a+3b+c
6=9a+3b+c

First we will subtract Equation #1 from Equation #2 in order to eliminate ‘c’:
3=4a+2b+c
1=a+b+c
The following equation (we will call it Equation #4)...