# SL Math Internal Assessment: Stellar Numbers

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Order NowPart 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 βnβ 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 βnβ in order to define a relationship between the term number and the term value. The constant second differences indicate that βnβ 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 βnβ 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 βnβ, we will use the general form: tn=an2+bn+c where tn is the term value (number of dots) and βnβ is the term number.

In order to determine the general statement of the sequence, we can simply solve for the values of the constants βaβ βbβ and βcβ. 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 βcβ:

3=4a+2b+c

1=a+b+c

The following equation (we will call it Equation #4) is obtained: 2=3a+b

The same will be done by subtracting Equation #2 from Equation #3:

6=9a+3b+c

3=4a+2b+c

The following equation (we will call it Equation #5) is obtained: 3=5a+b

Now that we have eliminated the constant βcβ from the equations, we will eliminate βbβ in order to solve for the value of βaβ by subtracting Equation #4 from Equation #5:

3=5a+b

2=3a+b

The following equation is obtained: 1=2a

Solve for βaβ: a=12

Now we can substitute the value of βaβ into Equation #4 to solve for βbβ:

2=312+b

b=12

Using the values of βaβ and βbβ and substituting them into Equation #1 we can solve for βcβ:

1=12+12+c

c=0

Now that we have defined the values for all 3 constants, we can conclude the general statement for the sequence by substituting the values into the primary quadratic equation:

General statement: tn=12n2+12n Factored form: tn=12n(n+1)

Test the general statement:

Term 5: t5=125[(5)+1] Term 8: t8=1288+1 The tests are valid. =15 =36

Therefore, the general statement that represents the nth triangular number is: tn=12n(n+1)

Part 2: Stellar shapes are defined with a βpβ number of vertices. Below is a series of 6-vertex stellar shapes. The numbers of dots in each stage are examples of p-stellar numbers. The first five representations for 6-stellar numbers (a star with 6 vertices) are shown in the five stages S1-S5 below:

S1 S2 S3 S4 S5

1 13 27 73 121 Term Number (n)| Number of Dots (Sn)| First Difference| Second Difference| 1| 1| -| -|

2| 13| 12| -|

3| 37| 24| 12|

4| 73| 36| 12|

5| 121| 48| 12|

6| 181| 60| 12|

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 12, the difference between the number of dots in sets 3 and 2 is 24, the difference between the number of dots in sets 4 and 3 is 36, and so on. The differences between each consecutive pair of 6-stellar numbers increase constantly by 12 each time. This dictates that there is a constant second difference of 12 for this sequence. When the number of dots for each term in the sequence is observed, no patterns can be clearly identified. However, by subtracting 1 from each of the 6-stellar numbers, the resulting numbers are found to be factors of 12. Also, by multiplying 12 by the sum of the n values from (n-1) to 0, and adding the 1 we had subtracted earlier, the number of dots in the sequence is obtained. Term Number (n)| Number of Dots (Sn)|

1| 0+1|

2| 12+1|

3| 36+1|

4| 72+1|

5| 120+1|

The new pattern can be analyzed and tested:

S1=12(0)+1 =1

S2=12(1)+1 =13

S3=12(2+1)+1 =37

S4=12(3+2+1)+1 =73

S5=12(4+3+2+1)+1 =121

Using this pattern, an expression for the 6-stellar number at stage S7 can be obtained:

S7=126+5+4+3+2+1+1

=1221+1

=253 The 6-stellar number at stage S7 is 253.

Just as we had induced in part 1 of this portfolio, the second difference indicates that the general statement for the sequence will be a quadratic in the form: Sn=an2+bn+c Just as we had done in part 1, in order to obtain the general statement for the 6-stellar number at stage Sn in terms of n, we can simply solve for it by a means of the elimination method in order to get the values for the constants βaβ βbβ and βcβ.

Equation #1 – For stage 1: S1=an2+bn+c

1=12a+1b+c

1=a+b+c

Equation #2 – For stage 2: S2=an2+bn+c

13=22a+2b+c

13=4a+2b+c

Equation #3 – For stage 3: S3=an2+bn+c

37=32a+3b+c

37=9a+3b+c

First we will subtract Equation #1 from Equation #2 in order to eliminate βcβ:

13=4a+2b+c

1=a+b+c

The following equation (we will call it Equation #4) is obtained: 12=3a+b

The same will be done by subtracting Equation #2 from Equation #3:

37=9a+3b+c

13=4a+2b+c

The following equation (we will call it Equation #5) is obtained: 24=5a+b

Now that we have eliminated the constant βcβ from the equations, we will eliminate βbβ in order to solve for the value of βaβ by subtracting Equation #4 from Equation #5:

24=5a+b

12=3a+b

The following equation is obtained: 12=2a

Solve for βaβ: a=6

Now we can substitute the value of βaβ into Equation #4 to solve for βbβ:

12=36+b

b=-6

Using the values of βaβ and βbβ and substituting them into Equation #1 we can solve for βcβ:

1=6+-6+c

c=1

Now that we have defined the values for all 3 constants, we can conclude the general statement for the sequence by substituting the values into the primary quadratic equation:

General statement: Sn=6n2-6n+1 Factored form: Sn=6nn-1+1

Test the general statement:

Term 4: S4=644-1+1 Term 5: S5=655-1+1 The tests are valid. =73 =121

Therefore, the general statement that represents the 6-stellar number at stage Sn: Sn=6nn-1+1

The first five representations for 5-stellar numbers (a star with 5 vertices) are shown in the five stages S1-S5 below: S1 S2 S3 S4

S5

1 11 31 61 101 Term Number (n)| Number of Dots (Sn)| First Difference| Second Difference| 1| 1| -| -|

2| 11| 10| -|

3| 31| 20| 10|

4| 61| 30| 10|

5| 101| 40| 10|

6| 151| 50| 10|

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 10, the difference between the number of dots in sets 3 and 2 is 20, the difference between the number of dots in sets 4 and 3 is 30, and so on. The differences between each consecutive pair of 5-stellar numbers increase constantly by 10 each time. This dictates that there is a constant second difference of 10 for this sequence. When the number of dots for each term in the sequence is observed, no patterns can be clearly identified. However, by subtracting 1 from each of the 5-stellar numbers, the resulting numbers are found to be factors of 10. Also, by multiplying 10 by the sum of the n values from (n-1) to 0, and adding the 1 we had subtracted earlier, the number of dots in the sequence is obtained. Term Number (n)| Number of Dots (Sn)|

1| 0+1|

2| 10+1|

3| 30+1|

4| 60+1|

5| 100+1|

The new pattern can be analyzed and tested:

S1=10(0)+1 =1

S2=10(1)+1 =11

S3=10(2+1)+1 =31

S4=10(3+2+1)+1 =61 S5=10(4+3+2+1)+1 =101

Using this pattern, an expression for the 5-stellar number at stage S7 can be obtained:

S7=106+5+4+3+2+1+1

=1021+1

=211 The 5-stellar number at stage S7 is 211.

Just as we had induced in part 1 of this portfolio, the second difference indicates that the general statement for the sequence will be a quadratic in the form: Sn=an2+bn+c Just as we had done in part 1, in order to obtain the general statement for the 6-stellar number at stage Sn in terms of n, we can simply solve for it by a means of the elimination method in order to get the values for the constants βaβ βbβ and βcβ.

Equation #1 – For stage 1: S1=an2+bn+c

1=12a+1b+c

1=a+b+c

Equation #2 – For stage 2: S2=an2+bn+c

11=22a+2b+c

11=4a+2b+c

Equation #3 – For stage 3: S3=an2+bn+c

31=32a+3b+c

31=9a+3b+c

First we will subtract Equation #1 from Equation #2 in order to eliminate βcβ:

11=4a+2b+c

1=a+b+c

The following equation (we will call it Equation #4) is obtained: 10=3a+b

The same will be done by subtracting Equation #2 from Equation #3:

31=9a+3b+c

11=4a+2b+c

The following equation (we will call it Equation #5) is obtained: 20=5a+b

Now that we have eliminated the constant βcβ from the equations, we will eliminate βbβ in order to solve for the value of βaβ by subtracting Equation #4 from Equation #5:

20=5a+b

10=3a+b

The following equation is obtained: 10=2a

Solve for βaβ: a=5

Now we can substitute the value of βaβ into Equation #4 to solve for βbβ:

10=35+b

b=-5

Using the values of βaβ and βbβ and substituting them into Equation #1 we can solve for βcβ:

1=5+-5+c

c=1

Now that we have defined the values for all 3 constants, we can conclude the general statement for the sequence by substituting the values into the primary quadratic equation:

General statement: Sn=5n2-5n+1 Factored form: Sn=5nn-1+1

Test the general statement:

Term 4: S4=544-1+1 Term 5: S5=555-1+1 The tests are valid. =61 =101

Therefore, the general statement that represents the 5-stellar number at stage Sn: Sn=5nn-1+1

The first five representations for 4-stellar numbers (a star with 4 vertices) are shown in the five stages S1-S5 below: S1 S2 S3 S4 S5

Term Number (n)| Number of Dots (Sn)| First Difference| Second Difference| 1| 1| -| -|

2| 9| 8| -|

3| 25| 16| 8|

4| 49| 24| 8|

5| 81| 32| 8|

6| 121| 40| 8|

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 8, the difference between the number of dots in sets 3 and 2 is 16, the difference between the number of dots in sets 4 and 3 is 24, and so on. The differences between each consecutive pair of 4-stellar numbers increase constantly by 8 each time. This dictates that there is a constant second difference of 8 for this sequence. When the number of dots for each term in the sequence is observed, no patterns can be clearly identified. However, by subtracting 1 from each of the 4-stellar numbers, the resulting numbers are found to be factors of 8. Also, by multiplying 8 by the sum of the n values from (n-1) to 0, and adding the 1 we had subtracted earlier, the number of dots in the sequence is obtained. Term Number (n)| Number of Dots (Sn)|

1| 0+1|

2| 8+1|

3| 24+1|

4| 48+1|

5| 80+1|

The new pattern can be analyzed and tested:

S1=8(0)+1 =1

S2=8(1)+1 =9

S3=8(2+1)+1 =25

S4=8(3+2+1)+1 =49 S5=8(4+3+2+1)+1 =81

Using this pattern, an expression for the 5-stellar number at stage S7 can be obtained:

S7=86+5+4+3+2+1+1

=821+1

=169 The 4-stellar number at stage S7 is 169.

Just as we had induced in part 1 of this portfolio, the second difference indicates that the general statement for the sequence will be a quadratic in the form: Sn=an2+bn+c Just as we had done in part 1, in order to obtain the general statement for the 6-stellar number at stage Sn in terms of n, we can simply solve for it by a means of the elimination method in order to get the values for the constants βaβ βbβ and βcβ.

Equation #1 – For stage 1: S1=an2+bn+c

1=12a+1b+c

1=a+b+c

Equation #2 – For stage 2: S2=an2+bn+c

9=22a+2b+c

9=4a+2b+c

Equation #3 – For stage 3: S3=an2+bn+c

25=32a+3b+c

25=9a+3b+c

First we will subtract Equation #1 from Equation #2 in order to eliminate βcβ:

9=4a+2b+c

1=a+b+c

The following equation (we will call it Equation #4) is obtained: 8=3a+b

The same will be done by subtracting Equation #2 from Equation #3:

25=9a+3b+c

9=4a+2b+c

The following equation (we will call it Equation #5) is obtained: 16=5a+b

Now that we have eliminated the constant βcβ from the equations, we will eliminate βbβ in order to solve for the value of βaβ by subtracting Equation #4 from Equation #5:

16=5a+b

8=3a+b

The following equation is obtained: 8=2a

Solve for βaβ: a=4

Now we can substitute the value of βaβ into Equation #4 to solve for βbβ:

8=34+b

b=-4

Using the values of βaβ and βbβ and substituting them into Equation #1 we can solve for βcβ:

1=4+-4+c

c=1

Now that we have defined the values for all 3 constants, we can conclude the general statement for the sequence by substituting the values into the

primary quadratic equation:

General statement: Sn=4n2-5n+1 Factored form: Sn=4nn-1+1

Test the general statement:

Term 4: S4=444-1+1 Term 5: S5=455-1+1 The tests are valid. =49 =81

Therefore, the general statement that represents the 4-stellar number at stage Sn: Sn=4nn-1+1

The first five representations for 7-stellar numbers (a star with 7 vertices) are shown in the five stages S1-S5 below: S1 S2 S3 S4 S5

Term Number (n)| Number of Dots (Sn)| First Difference| Second Difference| 1| 1| -| -|

2| 15| 14| -|

3| 43| 28| 14|

4| 85| 42| 14|

5| 141| 56| 14|

6| 211| 70| 14|

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 14, the difference between the number of dots in sets 3 and 2 is 28, the difference between the number of dots in sets 4 and 3 is 42, and so on. The differences between each consecutive pair of 7-stellar numbers increase constantly by 14 each time. This dictates that there is a constant second difference of 14 for this sequence. When the number of dots for each term in the sequence is observed, no patterns can be clearly identified. However, by subtracting 1 from each of the 7-stellar numbers, the resulting numbers are found to be factors of 14. Also, by multiplying 14 by the sum of the n values from (n-1) to 0, and adding the 1 we had subtracted earlier, the number of dots in the sequence is obtained. Term Number (n)| Number of Dots (Sn)|

1| 0+1|

2| 14+1|

3| 42+1|

4| 84+1|

5| 140+1|

The new pattern can be analyzed and tested:

S1=14(0)+1 =1

S2=14(1)+1 =15

S3=14(2+1)+1 =43

S4=14(3+2+1)+1 =85 S5=14(4+3+2+1)+1 =141

Using this pattern, an expression for the 7-stellar number at stage S7 can be obtained:

S7=146+5+4+3+2+1+1

=1421+1

=295 The 7-stellar number at stage S7 is 295.

Just as we had induced in part 1 of this portfolio, the second difference indicates that the general statement for the sequence will be a quadratic in the form: Sn=an2+bn+c Just as we had done in part 1, in order to obtain the general statement for the 6-stellar number at stage Sn in terms of n, we can simply solve for it by a means of the elimination method in order to get the values for the constants βaβ βbβ and βcβ.

Equation #1 – For stage 1: S1=an2+bn+c

1=12a+1b+c

1=a+b+c

Equation #2 – For stage 2: S2=an2+bn+c

15=22a+2b+c

15=4a+2b+c

Equation #3 – For stage 3: S3=an2+bn+c

43=32a+3b+c

43=9a+3b+c

First we will subtract Equation #1 from Equation #2 in order to eliminate βcβ:

15=4a+2b+c

1=a+b+c

The following equation (we will call it Equation #4) is obtained: 14=3a+b

The same will be done by subtracting Equation #2 from Equation #3:

43=9a+3b+c

15=4a+2b+c

The following equation (we will call it Equation #5) is obtained: 28=5a+b

Now that we have eliminated the constant βcβ from the equations, we will eliminate βbβ in order to solve for the value of βaβ by subtracting Equation #4 from Equation #5:

28=5a+b

14=3a+b

The following equation is obtained: 14=2a

Solve for βaβ: a=7

Now we can substitute the value of βaβ into Equation #4 to solve for βbβ:

14=37+b

b=-7

Using the values of βaβ and βbβ and substituting them into Equation #1 we can solve for βcβ:

1=7+-7+c

c=1

Now that we have defined the values for all 3 constants, we can conclude the general statement for the sequence by substituting the values into the primary quadratic equation:

General statement: Sn=7n2-7n+1 Factored form: Sn=7nn-1+1

Test the general statement:

Term 4: S4=744-1+1 Term 5: S5=755-1+1 The tests are valid. =85 =141

Therefore, the general statement that represents the 7-stellar number at stage Sn: Sn=7nn-1+1

There has been a distinct relationship between the βpβ values and their corresponding general statements for the p-stellar numbers. 5-Stellar: Sn=5nn-1+1

6-Stellar: Sn=6nn-1+1

7-Stellar: Sn=7nn-1+1

The coefficient in front of the βnβ in each general statement corresponds to the βpβ values. The rest of the equation however remains unchanged as the number of vertices (βpβ value) is changed. Therefore it can be induced that the general statement, in terms of βpβ and βnβ, that generates the sequence of p-stellar numbers for any value of βpβ at stage Sn is: Sn=pnn-1+1

Now to test it:

7-Stellar: S6=766-1+1 5-Stellar: S7=577-1+1

=211 =211 6-Stellar: S3=633-1+1 4-Stellar: S5=455-1+1 =37 =81 The tests are valid. The general statement that generates the sequence of p-stellar number for any value of βpβ at stage Sn is: Sn=pnn-1+1 However, there are restrictions related to finding p-stellar numbers. 1. The value of βpβ must be greater than 2

This is simply due to the fact that 2-stellar shapes (stars with 2 vertices) or 1-stellar shapes do not exist. It is impossible to draw a star with 2 or fewer vertices, therefore those values of p are inexistent and restrictions must be indicated. 2. βpβ must be an element of a real number

Imaginary numbers, such as -1 are not possible values for βpβ. They are inexistent and cannot be created into stellar shapes, meaning that there are no imaginary stellar numbers that can be solved for. 3. The value of βpβ cannot be rational

Stellar shapes are made of set numbers of vertices. Each vertex is counted as a single whole number. Rational numbers of vertices cannot be created due to it being impossible to have a fraction of a vertex. There are also restrictions regarding the term number (βnβ): 1. The term number must be of positive integer

Any sequence starts at 0. It is not possible to have a sequence which begins in the negative region due to it being impossible to a negative term. It does not exist, and will not produce significant information by determining its inexistent term value. 2. βnβ must be an element of a real number

Like it is mentioned above, imaginary numbers do not exist, and cannot be used as a term number. 3. The value of βnβ cannot be rational

For the same reason as to why there cannot be rational βpβ values, term numbers define a specific point in a sequence. Each term number has a corresponding term value. However, rational term numbers do not exist, and have no term value. So the restrictions on βpβ and βnβ are as follows:

3 β€ pβ€β nβR pβ Q 0 β€ nβ€β nβR nβ Q