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.^[1]

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...) ^[2]

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

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 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=  
Now that I know that the first part of the formula is n2 I can proceed to find the values of ‘b’ and ‘c’.

This second difference tells me the value for ‘b’ which is equal to . To find the value of ‘c’ I will use the previous methods:

Using n=4:

 n2+ n + c = 51
 (4)2+ (4) + c = 51
50 + c = 51
c = 1

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

Using n=1:

 n2+ n + 1 = 6
 (1)2+ (1) + 1 = 6

Using n=6:

 n2+ n + 1 = 106
 (6)2+ (6) + 1 = 106

Therefore the general statement for this shape is:

n2 +n + 1

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

Now, a p value of 6 will be used based on a regular hexagon:

Again, another pattern related to the pattern in other shapes has been detected. The relationship discovered is that the number of dots in the next stage is equal to the number of dots in the previous stage plus the term of the current stage multiplied by 6.  

I.e.:  6Sn = 6Sn-1+ 6n

To prove this equation an existing example from above will be used.
For Stage 2:
6Sn = 6Sn-1 + 6n
6S2= 6S1 + (6x2)
=7+ 12
=19

For Stage 4:
6Sn = 6Sn-1 + 6n
6S4= 6S3 + (6x4)
=37+ 24
=61

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 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=  = 3

Now that I know that the first part of the formula is 3n2 I can proceed to find the values of ‘b’ and ‘c’.

This second difference tells me the value for ‘b’ which is equal to 3. To find the value of ‘c’ I will use the previous methods:

Using n=4

3n2+3n + c = 61
3(4)2+3(4) + c = 61
60 + c = 61
c = 1

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

Using n=6:

3n2+3n + 1 = 127
3(6)2+3(6) + 1 = 127

Using n=2:

3n2+3n + c = 19
3(2)2+3(2) + 1 = 19

Therefore the general statement for this shape is:

3n2 +3n + 1

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

Finally, a p value of 8 will be analysed:

Once again, another pattern was observed as the stage numbers developed. This time, the connection discovered is that the number of dots in the next stage is equal to the number of dots in the previous stage plus the term of the current stage multiplied by 16.  

I.e.:  8Sn = 8Sn-1+ 16n

For Stage 3:
8Sn = 8Sn-1 + 16n
8S3= 6S2 + (16x3)
=49+ 48
=97

For Stage 6:
8Sn = 8Sn-1 + 16n
8S6= 8S5 + (16x6)
=241+ 96
=337

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 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=  = 8

Now that I know that the first part of the formula is 8n2 I can proceed to find the values of ‘b’ and ‘c’.

This second difference tells me the value for ‘b’ which is equal to 8. To find the value of ‘c’ I will use the previous methods:

Using n=4

8n2+8n + c = 161
8(4)2+8(4) + c = 161
160 + c = 161
c = 1

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

Using n=2:

8n2+8n + 1 = 49
8(4)2+8(4) + 1 = 49

Using n=3:

8n2+8n + 1 = 97
8(3)2+8(3) + 1 = 97

Therefore the general statement for this shape is:

8n2 +8n + 1

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

7. Hence, produce the general statement, in terms p and n, that generates the sequence of p-stellar numbers for any value of p at stage Sn.

When analyzing all the examples given, it is clear that there is no regular general pattern for the values of a, b and c that relate to the values of p and n. For polygons 2, 3 and 6 it seems that the values of a and b are the equal to the values of p. However, for polygons 4 and 5 it appears the values of a and b are half the value of p.

Nevertheless, I noticed that some shapes (such as polygon 2, 3 and 6) can be considered to have double their number of vertices, if we include the points that go in i.e. the concave lines. This arouses the question: what are vertices? A vertex should be considered “the common endpoint of two or more rays or line segments (…) Vertex typically means a corner or a point where lines meet.” ^[3] If we took this and followed it exactly then polygon 2 would have 12 vertices, instead of 6, polygon 3 would have 8, instead of 4, and polygon 6 would have 16 vertices, instead of 8. Using this these terms I can now find a general statement:

Polygon	pSn	a	b	c	General Statement
1	3Sn			1	...read more. Conclusion Accepting that all points at which lines meet as vertices is also important for the matter, as a stellar shape is assumed to be one that has the same number of concave and convex points. Another limitation is that the values in the sequence, the values of p and the values of n all have to be positive, as a negative value in each of these would be impossible to draw. Explain how you arrived at the general statement. Initially I attempted to see whether there was a relationship between the number of dots and the values of ‘a’ and ‘b’. Seen there was none, I decided to create a table so that the ideas were more organized. I then, started to see the connection. At first I did not consider the concave points in some of the polygons to be vertices; if we were strictly speaking they are p-pointed stars. However when I considered these to be vertices and altered my general table I could see the immediate relationship between the values of ‘a’ and ‘b’ and the value of p. Since ‘a’ was equal to ‘b’ in most cases it was simple to find the relationship. As a final point c was continuously 1 therefore I believe it to be 1 for the general expression derived.            [1]All the questions in dark blue are from the Oporto British School Maths internal Assessment handout 2011 [2] Steps followed by “The nth term of quadratics” at http://www.pearsonpublishing.co.uk/education/samples/S_492153.pdf [3]. Definition from http://www.mathopenref.com/vertex.html ...read more. This student written piece of work is one of many that can be found in our International Baccalaureate Maths section. Found what you're looking for? Start learning 29% faster today 150,000+ documents available Just £6.99 a month Get Full Access Now or Learn more Not the one? Search for your essay title...Search Join over 1.2 million students every month Accelerate your learning by 29% Unlimited access from just £6.99 per month See related essays Related International Baccalaureate Maths essays Math IB HL math portfolio type II. Deduce the formula Sn = ... = Sk + (S3k - S2k) 2(q - p) = p + (S3k - q) 2q - 2p = p + S3k - q S3k = 3q - 3p S0 Sk S2k S3k S4k p q - p q 3q - 3p 2(S3k - S2k) Maths Internal Assessment -triangular and stellar numbers Question 5: Find a general statement for the 6-stellar number at stage Sn in terms of n. Similar to using square numbers to find out triangular numbers, finding the general statement for the 6-stellar number will require the manipulation of triangular numbers. Stellar numbers The final general statement for the triangle will be expressed in terms of n: if the general statement for the nth triangle were multiplied all together then the equation would become the quadratic function: . In order to test for validity the graphing program "Graph 4.3" was used to demonstrate the validity of the general statement. Stellar numbers. The aim of the current investigation is to consider different geometric ... 3 n = 3 u3 = 6 u3 = u2 + n = 6 n = 4 u4 = 10 u4 = u3 + n = 10 n = 5 u5 = 15 u5 = u4 + n = 15 n = 6 u6 = 21 u6 = u5 + Stellar Numbers. After establishing the general formula for the triangular numbers, stellar (star) shapes ... + c 6 = 9a + 3b + c Thereafter polysmlt on the graphic display calculator is used in order to retrieve the values of a, b, and c. The coefficients above are simply plugged into the calculator; this can be seen in the screenshots below Hence the general formula Stellar Numbers Investigation Portfolio. 2P] Sn = 1 + [(n-1) �2] x [2 x 12 + (n - 2) 12] S7 = 1 + [(7-1) �2] x [2 x 12 + (7- 2) 12] S7 = 1 + [(6) �2] x [24 + (5) How many pieces? In this study, the maximum number of parts obtained by n ... R1 = 2 --> 1 + 1 --> R0 + S0 R2 = 4 --> 2 + 2 --> R1 + S1 R3 = 7 --> 4 + 3 --> R2 + S2 R4 = 11 --> 7 + 4 --> R3 + S3 R5 = 16 --> 11 + Stellar Numbers Portfolio. The simplest example of these is square numbers, but over the ... After looking at the diagrams above the following information can be easily determined. S1=1 S2=13 S3=37 S4=73 From this information, the pattern that I determined in that each term is 12n more than the previous term, in which n is equal to the term number of the previous term. See more essays Over 160,000 pieces of student written work Annotated by experienced teachers Ideas and feedback to improve your own work Get full access now Short of cash - and want FREE access? Submit one of your essays and get a FREE DAY PASS. You don't need to use your pass straight away - you can save it for later! Click here to submit your essay Save Sign up now Want to read the rest? Sign up to view the whole essay and download the PDF for anytime access on your computer, tablet or smartphone. Read more (The above preview is unformatted text) Found what you're looking for? Start learning 29% faster today 150,000+ documents available Just £6.99 a month Get Full Access Now or Learn more Looking for expert help with your Maths work? Check out our FREE Study Guides: Created by teachers, our study guides highlight the really important stuff you need to know. Take me to free Study Guides Or get inspiration from these FREE essays: pH titration curves Lab Report. How does the use of a strong acid ... China One Child Policy 4 star(s) Don't have an account yet? Create one now! Already have an account? Log in now!