Introduction: The aim in this task is to consider geometric shapes which lead to special numbers. The simplest example of these are square numbers, 1, 4, 9, 16, which can be represented by squares of side 1, 2, 3 and 4.

Task 1The first task is to complete the triangular numbers sequence with three more terms. Here are the initial terms:

1 3 6 10 15

By looking at this sequence, we can create a picture of what the pattern is. From the first term to the second term, there is an increase of 2, from the second to third is an increase of 3, and from the third to the fourth is an increase of 4 etc.

The pattern is as follows:

0+1=11+2=33+3=66+4=1010+5=15And from here it continues as:15+6=2121+7=2828+8=36

Task 2The second task is to find a general statement that represents the nth triangular number in terms of n.

Based on the pattern above, we can see that it is a sequence and not a series but also not an arithmetic sequence since there is no fixed number being added (common difference) to each number and it is not a geometric sequence because each number is not being multiplied by the same constant (common ratio). Though it is neither geometric nor arithmetic, there is still a pattern. The numbers added to each consecutive term go through the number line: 1+2, 3+3, 6+4, 10+5 and so on. To create a general statement for the nth number, we need sequence to be arithmetic or geometric. Here the ratio comes into place:

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nx

where n is the nt h term and x is the number of the nt h term. This is puts the pattern in a ratio.

nx=nx

nx=1

1

n=x

Using this as a base, we can repeat it for the rest of the terms in the sequence:nx=2

3∙nx=3

6∙nx= 4

10∙nx= 5

15∙nx= 6

21∙nx= 7

28∙nx= 8

36

Note that the bolded numbers are the three extra terms.

When simplified, the sequence is: n ,32n ,2n ,

52n ,3n ,

72n ,4 n ,

92n

This is now in the format of an arithmetic sequence since one half is being added to each consecutive term. We can now use the arithmetic sequence’s formula for finding the nth term which is: un=u1+ (n−1 )d

We then substitute in what values we know to create the general statement.

When u1=n and d=n2

so the general statement is un=u1+ (n−1 ) n2

To confirm this, test with n=4and un=10:

10=4+(4−1) 42

1 0=4+(3) 42

10=4+610=10 hence this general statement is valid.

The general statement ultimately can be simplified to n2+n2

So it should the sequence should look as follows:

1 3 6 10 15 21 28 36

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Task 3Task 3 is based on the stellar shapes with p vertices, leading to p-stellar numbers. In this task we have to find the number of dots in each stage up to s6.The initial terms are:

The number of dots in each term is as follows:s1=1s2=13s3=37s4=73s5=121s6=181

The patter is as follows:1+0=11+12=1313+24=3737+36=7373+48=121121+60=181Note that bold are the terms added.

Based on this pattern, each term is found by the previous term being added to by multiples of 12 starting at 0: 1+0=1 ,1+12=13 ,13+24=37∧37+36=73. However the multiple is one less than the n of each term. For example when n=3, 12×2 when n=4 ,12×3. The pattern starts initially with 1 before anything is added to it.

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Task 4In task 4 we have to find an expression for the 6-stellar number at stage s7.

This can be done using the patterns found in the previous task so the expression is:s7=s6+12 (n−1 )Which would become when s6=181∧n=7s7=181+12 (7−1 )s7=181+12 (6 )s7=181+72s7=253

Task 5The fifth task is to find a general statement for the 6-stellar number at stage sn in terms of n.

Using the general statement from task 2 which is: n(n+1)

2Using this statement, we have to modify it so that it suits the needs of the 6-stellar number by using the patterns found in task 3, one less than the nth and the extra one. We need to do so since

the nx

ratio does not apply here. This would then lead to n=12n since each term is multiplied by

the multiples of 12, (n+1) to (n−1) since each of the numbers that 12 is multiplied to is one less than the n of the term, and the additional 1 since the pattern starts initially with 1 before anything is added to it. This then results in the task 3 general statement to become: 12n (n−1)

2+1=6n (n−1 )+1. This is the general statement for the 6-stellar number.

Task 6The sixth task is to repeat the steps above for other values of p where p is the number of vertices on the star. For example the 6-stellar has a p=6 and a 5-stellar has a p=5.I will be repeating the steps for 4-stellar and 5-stellar.

For p=4 or 4-stellar numbers, these are the values to s4:

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s1 s2 s3 s4

The number of dots in each is as follows:s1=1s2=9s3=25s4=49s5=81s6=121

The pattern is as follows:1+0=11+8=99+16=2525+24=4949+32=8181+40=121The bolded terms are not present in the diagram.

In this sequence, we can see that each term is found by adding a multiple of 8 to the previous term, the multiple is one less than the n of the term and the sequence starts initially with 1 before anything is added to it. Using this pattern, we can construct an expression for s7 which is: s7=s6+8(n−1) which would then become when s6=121, n=7:s7=121+8(7−1)s7=121+8(6)s7=121+48s7=169From this we can get the general statement by adjusting the statement we gained from task 2, n(n+1)

2.

Here we need to replace n with 8n since each term is multiplied by multiples of 8, (n+1) to(n−1) since each of the numbers that 8 is multiplied is one less than the n of the term, and the

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additional 1 since the pattern starts initially with 1 before anything is added to it. This will result

in the general statement to become: 8n(n−1)

2+1=4n (n−1 )+1.

The sequence should now look as follows:

s1 s2 s3 s4 s5 s6

To test the validity of this statement, use n=3 where n is the term number.

sn=4n (n−1 )+1s3=4 (3)(3−1 )+1s3=12 (2 )+1s3=24+1s3=25This general statement is valid

For p=5 or 5-stellar numbers, these are the values to s4:

s1 s2 s3 s4 The number of dots in each is as follows:s1=1s2=11s3=31s4=61s5=101

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s6=151

The pattern is as follows:1+0=11+10=1111+20=3131+30=6161+40=101101+50=151The bolded terms are not present in the diagram.

In this sequence, we can see that each term is found by adding a multiple of 10 to the previous term, the multiple is one less than the n of the term and the sequence starts initially with 1 before anything is added to it. Using this pattern, we can construct an expression for s7 which is: s7=s6+10(n−1) which would then become when s6=151, n=7:s7=151+10 (7−1)s7=151+10 (6)s7=151+60s7=211From this we can get the general statement by adjusting the statement we gained from task 2, n(n+1)

2.

Here we need to replace n with 10n since each term is multiplied by multiples of 10, (n+1) to(n−1) since each of the numbers that 10 is multiplied is one less than the n of the term, and the additional 1 since the pattern starts initially with 1 before anything is added to it. This will result

in the general statement to become: 10n(n−1)

2+1=5n (n−1 )+1.

The sequence should now look as follows:

s1 s2 s3 s4 s5 s6

To test the validity of this statement, use n=3 where n is the term number.

sn=5n (n−1 )+1s3=5 (3)(3−1 )+1

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s3=15 (2 )+1s3=30+1s3=31This general statement is valid

Another way to get a general statement is through means of technology. I will be using a calculator.Calculator used: TI-84 Plus Silver EditionFrom the following table, enter n into List 1 and Sn into List 2n 1 2 3 4Sn 1 11 31 61

After, use the calculation function to calculate the Quadratic regression using List 1 and List 2.It displaysQuadRegy=a x2+bx+ca=5b=−5c=1Substitute in the known values for a ,band c and substitute n for x and Sn for y.Sn=5n2−5n+1

To test the validity, use n=4S4=5(4)2−5(4)+1S4=5(16)−20+1S4=80−20+1S4=61

This general statement is valid.

Task 7The seventh task is to produce a general statement, in terms of p and n, that generates the sequence of p-stellar numbers for any value of p at stage Sn.

When looking at all the general statements for 4-stellar, 5-stellar, and 6-stellar at Sn in terms on n, they are 4 n (n−1 )+1, 5n (n−1 )+1, and 6n (n−1 )+1 respectivley. One can see that there is a pattern and they are almost identical except for the first term however all the first terms are equal to each of their p values respectively so therefore we can say that the general statement in terms of p and n is: Sn=p (n ) (n−1 )+1.

Task 8The next task is to test the validity of the general statement.

To do so we can use the next stellar shape.

To test the validity, use p=7 and n=3

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Sn=p (n ) (n−1 )+1S3=7 (3 ) (3−1 )+1S3=21 (2 )+1S3=42+1S3=43

Since S1 must be equal to 1, we can use the general statement for 7-stellar to see if the above is true.sn=7n (n−1 )+1s3=7 (3 ) (3−1 )+1s3=21 (2 )+1s3=42+1s3=43

The general statement in terms of p and n is valid.

Task 9Discuss the scope or limitations of the general statement.

The test of the validity in Task 8 shows that the scope of the general statement lies within all stellar numbers, or stars. The p value given the fact that it is the number of vertices cannot be a fraction nor can it be a negative number. The Sn also cannot be a fraction or a negative number because it is the total number of dots and you cannot have a fraction of a dot or a negative dot nor can the n since it is the term number.

Task 10Explain how you arrived at the general statement.

I arrived at the general statement by first looking at the other general statements for each individual stellar number and finding the relationship between them. I saw that they were all identical except for the first term i.e. 4 n (n−1 )+1, 5n (n−1 )+1, and 6n (n−1 )+1 so to make a general statement in terms of p and n, I just replaced the number with p since I realized that the numbers were the same as the number of vertices the star had.

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