Stellar Numbers. In this task geometric shapes which lead to special numbers will be considered.

Maths Internal Assessment

Type 1 – Mathematical Investigation

Mathematics Standard Level

Stellar Numbers

September 2011

Luís Ferreira

Teacher: Mr.Robson

Aim – In this task geometric shapes which lead to special numbers will be considered.

For example the easiest of these are square numbers which can be represented by squares of side 1, 2, 3 and 4.

1. The following diagrams show a triangular pattern of evenly spaced dots. The numbers of dots in each diagram are examples of triangular numbers. Complete the triangular numbers sequence with three more terms. Find a general statement that represents the nth triangular number in terms of n.

Finding a general statement:

Now that three more terms have been drawn, the general statement can be found for the sequence; 1, 3, 6, 10, 15, 21, 28, 36.

To do this, the constant difference in the sequence will need to be found, as shown below. This is needed to determine the type of equation (linear, quadratic, cubic etc...)

The standard rules to find the general statement were researched and the following the method was put into practice for all the shapes in this portfolio.

If the second difference is a constant, the formula for the nth term contains n2 as in a quadratic formula i.e. ax2 + bx + c.

The value of ‘a’ is half the constant difference. In this example a=

Hence, that the first part of the formula is  n2. To find the rest of the formula, the differences between the values in the sequence and the values of n2 will need to be calculated.

This second difference illustrates the value for ‘b’ which is equal to.

However the value of ‘c’ has not yet been determined. It was calculated using an example:

Using n=2:

n2 +  n + c = 6
(2)2 +  (2) + c = 6
5 + c=6
c=1

From the example we can verify that ‘c’ must be equal to 1 to reach the desired figure.

To check that these are the correct values two more examples were used:

Using n=5
n2 +  n + 1 = 21
(5)2 +  (5) + 1 = 21

Using n=7:
n2 +  n + 1 = 36
(7)2 +  (7) + 1 = 36

With proof that the formula works it is concluded that the general statement for this pattern is:

n2 + n + 1

1. Consider stellar (star) shapes with p vertices, leading to p-stellar numbers. The first four representations for a star with six vertices are shown in the four stages S1-S4. The 6-stellar number at each stage is the total number of dots in the diagram.
2. Find the number of dots (i.e. the stellar number) in each stage up to S6. Organize the data so that you can recognize and describe any patterns.

1. Find an expression for 6-stellar number at stage S7

By observing the pattern in the table I can use it to calculate the number of dots at stage 7. If I add 12x6 to the previous number (253) I can calculate the number of dots in stage 7.

6S7 = 253 + (12x6)
= 337

A relationship was established in the sequence. This was determined by observing the sides of the shape and analyzing the table. There is a link between the number of sides and the number of dots on the side. There are 12 sides, each with 7 dots respectively. In fact, the number of dots in the next stage is equal to the number of dots in the previous stage, plus the multiplication of 12 and the term number.

I.e.:  6Sn = 6Sn-1 + 12n

To prove my equation I will use an existing example from above and then prove it with one stage further.

For Stage 3:
6Sn = 6Sn-1 + 12n
6S4= 6S3 + (12x3)
=
37 + 36
=73

For stage 8:
6Sn = 6Sn-1 + 12n
6S8 = 6S7 + (12x7)
=337 + 84
=433

6S8

With this formula proven I can reach a statement for stage 7:

6S7 = 6S7-1 + 12(7)

1. Find a general statement for the 6-stellar at stage Sn in terms of n

Finding a general statement:

Now that I have more terms and now that they have been drawn I can now find a general statement for the sequence, using the previous methods.

Again I will need to find the constant difference in the sequence in order to establish the type of equation.

Again the second difference is the constant therefore the formula for the nth term contains n2 as in the quadratic equation: ax2 + bx + c

The value of ‘a’ is half the constant difference. In this example a=  = 6

Now that I know that the first part of the formula is 6n2 I can proceed to find the values of ‘b’ and ‘c’ just as in the first step.

This second difference tells me the value for ‘b’ which is equal to 6.

Yet again we still have not determined what the value for ‘c’ is. So an example should help me find out

Using n=4:

6n2 +6n + c = 121
6(4)
2 +6(4) + c = 121
120 + c= 121
c= 1

To check that these are the correct values I used two more examples:

Using n=2
6n
2 +6n + c = 121·
6(2)
2 +6(2) + 1 = 121

Using n=6:
6n
2 +6n + 1 = 121
6(6)
2 +6(6) + 1 = 121

With enough proof that the general statement works I conclude that it is:

6n2 +6n + 1

As this is a quadratic equation, a graph was plotted to demonstrate how it expanded

1. Now repeat the steps for other values of p

Firstly, a shape with a p value of 4 will be considered:

Again, another pattern related to ...