→ Using n=2
6n2 +6n + c = 121·
6(2)2 +6(2) + 1 = 121
→ Using n=6:
6n2 +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
 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 the pattern in step 4 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 8.
I.e.: 4Sn = 4Sn1 + 8n
To prove this equation an existing example from above will be used.
For Stage 3:
4Sn = 4Sn1 + 8n
4S3= 4S2 + (8x3)
= 25 + 24
= 49
For Stage 5:
4Sn = 4Sn1 + 8n
4S5= 8S4 + (8x5)
= 81 + 40
=121
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 becomes 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= =4
Now that I know that the first part of the formula is 4n2 I can proceed to find the values of ‘b’ and ‘c’.
This second difference tells me the value for ‘b’ which is equal to 4. To find the value of ‘c’ I will use the previous methods:
Using n=2:
4n2 +4n + c = 25
4(2)2 +4(2) + c = 25
24 + c = 25
c = 1
To check that these are the correct values I used two more examples:
Using n=6
4n2 +4n + 1 = 169
4(6)2 +4(6) + 1 = 169
Using n=5:
4n2 +4n + 1 = 121
4(5)2 +4(5) + 1 = 121
Therefore the general statement for this shape is:
4n2 +4n + 1
As this is a quadratic equation, a graph was plotted to demonstrate how it expanded
This time, a shape with a p value of 5 (i.e. a 5vertices shape), will be shown:
Again, another ‘side’ association related. 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 5.
I.e.: 5Sn = 5Sn1 + 5n
For Stage 2:
5Sn = 5Sn1 + 5n
5S2 = 5S1 + (5x2)
= 6 + 10
=16
For stage 4:
5Sn = 5Sn1 + 5n
5S4 = 5S3 + (5x4)
=31 + 20
=51
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=
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 = 6Sn1 + 6n
To prove this equation an existing example from above will be used.
For Stage 2:
6Sn = 6Sn1 + 6n
6S2= 6S1 + (6x2)
= 7 + 12
=19
For Stage 4:
6Sn = 6Sn1 + 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 = 8Sn1 + 16n
For Stage 3:
8Sn = 8Sn1 + 16n
8S3= 6S2 + (16x3)
= 49 + 48
=97
For Stage 6:
8Sn = 8Sn1 + 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:

Hence, produce the general statement, in terms p and n, that generates the sequence of pstellar 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 or (…) Vertex typically means a corner or a point where lines meet.” ^{} 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:
Green indicates the modified terms according to the definition of a vertex.
It is plainly indicated that the ones that changed were the stellar (star) shapes as they are the ones with concave and convex lines.
It can clearly be seen that in almost all the cases ‘a’ and ‘b’ are half the values of p. Therefore for these types of shapes I can conclude that a = and that b =. All the way through, c is constantly equal to 1. All the general statements that were derived were a quadratic one, consequently, so is the general rule, hence the general rule for the shapes is:
pSn = n2+ n + 1
Another extra statement that can be formulated relates to the ‘side’ equations found out.
If we take the exact definition of a vertex as before we can consider another expression to find the nth term:
pSn = pSn1 + pn
Polygon 1 does not fit into any of the rules and this will be discussed in step 9.
 Test the validity of the statement.
Let’s firstly consider this general rule above and confirm it.
A shape of p value of 7 was created:
Once again, another pattern was observed as the stage numbers developed. This time, the link that was found was 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 7
I.e.: 7Sn = 7Sn1 + 7n
For Stage 5:
7Sn = 7Sn1 + 7n
7S5 = 7S4 + (7x5)
= 71 + 35
=106
For Stage 6:
7Sn = 7Sn1 + 7n
7S4 = 7S5 + (7x6)
= 106 + 42
=148
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=
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 = 71
(4)2 + (4) + c = 71
70 + c = 71
c = 1
To check that these are the correct value, two more examples were used:
Using n=1:
n2 + n + 1= 8
(1)2 + (1) + c = 8
Using n=6:
n2 + n + 1 = 148
(6)2 + (6) + c = 148
Therefore the general statement for this shape is:
n2 + n + 1
This proves that my general equation works. But, even though the general statement has been verified with 5 different shapes and an extra one, it does not apply to all shapes. For example for polygon 1 the rule does not apply, therefore there must be exceptions to the rule which derive other general equations. Let’s explore those exceptions.
The first thing investigated was how the rule changed when the line length increased by a factor more than 1.
Let’s consider the following shape with a p value of 4:
We can clearly observe that there is a difference in this shape. The subsequent stage increases the line length by 2 points instead of 1. Thus, the general statement was calculated for this shape to see if there was a difference between it and the one set in step 7.
Once again, another pattern was observed as the stage numbers developed.
Using the second rule in step 7 (pSn = pSn1 + pn) this formula should be:
I.e.: 4Sn = 4Sn1 + 8n
Notice that this time we are multiplying the current term by 8. In the last shapes it would be equal to p but this time we are multiplying by 2p therefore we can see the effect of changing the length of the line by 2 on this formula.
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= = 4
Now that the first part of the formula is 4n2 the values of ‘b’ and ‘c’ can be found.
This second difference tells me the value for ‘b’ which is equal to 4. To find the value of ‘c’ I will use the previous methods:
Using n=2
4n2 +4n + c = 25
4(2)2 +4(2) + c = 25
24 + c = 25
c = 1
To check that these are the correct values, two more examples were used:
Using n=3:
4n2 +4n + 1 = 49
4(2)2 +4(2) + 1 = 49
Using n=4:
4n2 +4n + 1 = 81
4(4)2 +4(4) + 1 = 81
According to the general rule formulated in question 7, the general statement should be 2n2+2n+1.
Yet, the general statement for this shape is:
4n2 +4n + 1
Therefore an increase in line length along the sequence by more than one point will prove to generate a new formula where a and b are both equal to p.
Another factor was considered: what if the number of dots in the centre of the shape varied? An extra shape was created to find out what could be the effect on the general rule:
Considering another p value of 4:
It was observed that this time a ‘side’ formula could not be generated as clearly for this shape as with the other ones.
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=1
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 2.
To find the value of ‘c’ I will use the previous methods:
Using n=5:
n2 +2n + c = 36
(5)2 +2(5) + c = 36
35 + c = 36
c = 1
To check that these are the correct values I used two more examples:
→ Using n=1
n2 +2n + c = 4
(1)2 +2(1) + 1 = 4
→ Using n=3:
n2 +2n + c = 16
(3)2 +2(3) + 1 = 36
With enough proof that the general statement works I conclude that it is:
n2 +2n + 1
According to the general rule formulated in question 7, the general statement for this case should be 2n2+2n+1. However this time the central dot did not stay consistent. A central dot did not exist when n would equal an odd number. There seems to be a pattern with odd numbers in this shape since on its first table this was underlined as well. Not having a consistent central dot also shows that a ‘side’ formula cannot be developed like the other polygons. Another issue applied here is that the in the sequence, the next level does not enclose the previous shape, in other words, the previous shape does not fit into the next shape like all the other.
Lastly, a polygon similar to polygon 1 was drawn, but, since polygon had no consistent central dot, one was drawn that always had a dot in the middle. Therefore a shape with a p value of 3 was drawn:
Once again, another pattern was observed as the stage numbers developed.
Using the second rule in step 7 (pSn = pSn1 + pn) this formula should be:
I.e.: 3Sn = 3Sn1 + 3n
Let’s verify this:
For Stage 3:
3Sn = 3Sn1 + 3n
3S3 = 3S2 + (3x3)
= 10 + 9
=19
For Stage 6:
3Sn = 3Sn1 + 3n
3S6 = 3S5 + (3x6)
= 46 + 18
=64
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=
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=2:
n2 + n + c = 10
(2)2 + (2) + c = 10
9 + c = 10
c = 1
To check that these are the correct values, two more examples were used:
Using n=3:
n2 + n + 1 = 19
(3)2 + (3) + 1 = 19
Using n=5:
n2 + n + 1 = 46
(5)2 + (5) + 1 = 46
Therefore the general statement for this shape is:
n2 + n + 1
This time, the rule generated in step 7 does apply to a shape with 3 vertices. This evidently shows the effect of having a constant central dot. Changing this factor seems to have changed the value of b, given that in polygon 1 it was .
 Discuss the scopes or limitations of the general statement.
When reviewing the general statement, some exceptions can obviously be found. Many factors affect values in the general statement of the shapes. Increasing the length line by more than 1, having a number other than one as central dots, having altering central dots, having none at all and seeing whether the shapes fit on top of each other are all aspects that are needed to be taken into account when formulating the overall expression. Therefore this statement will only work for shapes that have a constant central point, increase the length line by 1, and the previous shape must enclose inside the next one.
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 ppointed 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.
All the questions in dark blue are from the Oporto British School Maths internal Assessment handout 2011
Steps followed by “The nth term of quadratics” at http://www.pearsonpublishing.co.uk/education/samples/S_492153.pdf
. Definition from http://www.mathopenref.com/vertex.html