SL Math Internal Assessment: Stellar Numbers Sample Essay

Part 1: Below is a series of trigon patterned sets of points. The Numberss of points in each diagram are illustrations of triangular Numberss. Let the variable ‘n’ represent the term figure 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 ( Tennessee ) | 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 turning addition in the differences between each back-to-back set of Numberss of points. The difference between the figure of points in sets 2 and 1 is 2. the difference between the figure of points in sets 3 and 2 is 3. the difference between the figure of points in sets 4 and 3 is 4. and so on. The differences between each back-to-back brace of triangular Numberss increase invariably by 1 each clip. This dictates that there is a changeless 2nd difference of 1 for this sequence. The term values. or in this instance the figure of points. increase neither arithmetically nor geometrically.






This piece of information. shows that the general statement will necessitate an advocate on the variable ‘n’ in order to specify a relationship between the term figure and the term value. The changeless 2nd differences indicate that ‘n’ will be raised to the advocate 2. Differences can be seen to stand for derived functions. The 2nd difference represents the 2nd derived function. where the advocate on the primary variable has been reduced by 2. holding been differentiated twice. In the same mode. the first difference represents the first derived function where the advocate on the primary variable has been reduced by merely 1. In order to return to the primary equation. the advocate from the 2nd derivative equation will hold to be raised twice. which is why the advocate on ‘n’ will be 2 and the general statement will be in the signifier of a quadratic.

The general signifier for a quadratic is: y=ax2+bx+c
This being a general statement for a sequence. and our primary variable being ‘n’ . we will utilize the general signifier: tn=an2+bn+c where Tennessee is the term value ( figure of points ) and ‘n’ is the term figure.

In order to find the general statement of the sequence. we can merely work out for the values of the invariables ‘a’ ‘b’ and ‘c’ . In order to make so. we will necessitate to specify 3 equations and utilize the method of riddance to work out for the general statement. Elimination requires adding or deducting 2 equations in the purpose of extinguishing a constant/variable by first fixing them to portion the same coefficients with either the same or opposite whole numbers ( depending on whether you will be adding or deducting 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 deduct Equation # 1 from Equation # 2 in order to extinguish ‘c’ :
3=4a+2b+c
1=a+b+c
The undermentioned equation ( we will name it Equation # 4 ) is obtained: 2=3a+b
The same will be done by deducting Equation # 2 from Equation # 3:
6=9a+3b+c
3=4a+2b+c
The undermentioned equation ( we will name it Equation # 5 ) is obtained: 3=5a+b
Now that we have eliminated the changeless ‘c’ from the equations. we will extinguish ‘b’ in order to work out for the value of ‘a’ by deducting Equation # 4 from Equation # 5:
3=5a+b
2=3a+b









The undermentioned equation is obtained: 1=2a
Solve for ‘a’ : a=12
Now we can replace the value of ‘a’ into Equation # 4 to work out for ‘b’ :
2=312+b
b=12
Using the values of ‘a’ and ‘b’ and replacing them into Equation # 1 we can work out for ‘c’ :
1=12+12+c
c=0
Now that we have defined the values for all 3 invariables. we can reason the general statement for the sequence by replacing the values into the primary quadratic equation:
General statement: tn=12n2+12n Factored signifier: tn=12n ( n+1 )
Test the general statement:
Term 5: t5=125 [ ( 5 ) +1 ] Term 8: t8=1288+1 The trials are valid. =15 =36
Therefore. the general statement that represents the n-th triangular figure is: tn=12n ( n+1 )











Part 2: Leading forms are defined with a ‘p’ figure of vertices. Below is a series of 6-vertex leading forms. The Numberss of points in each phase are illustrations of p-stellar Numberss. The first five representations for 6-stellar Numberss ( a star with 6 vertices ) are shown in the five phases 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 turning addition in the differences between each back-to-back set of Numberss of points. The difference between the figure of points in sets 2 and 1 is 12. the difference between the figure of points in sets 3 and 2 is 24. the difference between the figure of points in sets 4 and 3 is 36. and so on. The differences between each back-to-back brace of 6-stellar Numberss increase invariably by 12 each clip. This dictates that there is a changeless 2nd difference of 12 for this sequence. When the figure of points for each term in the sequence is observed. no forms can be clearly identified. However. by deducting 1 from each of the 6-stellar Numberss. the resulting Numberss are found to be factors of 12. Besides. by multiplying 12 by the amount of the n values from ( n-1 ) to 0. and adding the 1 we had subtracted earlier. the figure of points 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 form 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 form. an look for the 6-stellar figure at phase S7 can be obtained:
S7=126+5+4+3+2+1+1
=1221+1
=253 The 6-stellar figure at phase S7 is 253.













Merely as we had induced in portion 1 of this portfolio. the 2nd difference indicates that the general statement for the sequence will be a quadratic in the signifier: Sn=an2+bn+c Merely as we had done in portion 1. in order to obtain the general statement for the 6-stellar figure at phase Sn in footings of n. we can merely work out for it by a agency of the riddance method in order to acquire the values for the invariables ‘a’ ‘b’ and ‘c’ .

Equation # 1 – For phase 1: S1=an2+bn+c
1=12a+1b+c
1=a+b+c
Equation # 2 – For phase 2: S2=an2+bn+c
13=22a+2b+c
13=4a+2b+c
Equation # 3 – For phase 3: S3=an2+bn+c
37=32a+3b+c
37=9a+3b+c







First we will deduct Equation # 1 from Equation # 2 in order to extinguish ‘c’ :
13=4a+2b+c
1=a+b+c
The undermentioned equation ( we will name it Equation # 4 ) is obtained: 12=3a+b
The same will be done by deducting Equation # 2 from Equation # 3:
37=9a+3b+c
13=4a+2b+c
The undermentioned equation ( we will name it Equation # 5 ) is obtained: 24=5a+b
Now that we have eliminated the changeless ‘c’ from the equations. we will extinguish ‘b’ in order to work out for the value of ‘a’ by deducting Equation # 4 from Equation # 5:
24=5a+b
12=3a+b
The undermentioned equation is obtained: 12=2a
Solve for ‘a’ : a=6
Now we can replace the value of ‘a’ into Equation # 4 to work out for ‘b’ :
12=36+b
b=-6
Using the values of ‘a’ and ‘b’ and replacing them into Equation # 1 we can work out for ‘c’ :
1=6+-6+c
c=1

















Now that we have defined the values for all 3 invariables. we can reason the general statement for the sequence by replacing the values into the primary quadratic equation:
General statement: Sn=6n2-6n+1 Factored signifier: Sn=6nn-1+1
Test the general statement:
Term 4: S4=644-1+1 Term 5: S5=655-1+1 The trials are valid. =73 =121
Therefore. the general statement that represents the 6-stellar figure at phase Tin: Sn=6nn-1+1
The first five representations for 5-stellar Numberss ( a star with 5 vertices ) are shown in the five phases 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 turning addition in the differences between each back-to-back set of Numberss of points. The difference between the figure of points in sets 2 and 1 is 10. the difference between the figure of points in sets 3 and 2 is 20. the difference between the figure of points in sets 4 and 3 is 30. and so on. The differences between each back-to-back brace of 5-stellar Numberss increase invariably by 10 each clip. This dictates that there is a changeless 2nd difference of 10 for this sequence. When the figure of points for each term in the sequence is observed. no forms can be clearly identified. However. by deducting 1 from each of the 5-stellar Numberss. the resulting Numberss are found to be factors of 10. Besides. by multiplying 10 by the amount of the n values from ( n-1 ) to 0. and adding the 1 we had subtracted earlier. the figure of points 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 form 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 form. an look for the 5-stellar figure at phase S7 can be obtained:
S7=106+5+4+3+2+1+1
=1021+1
=211 The 5-stellar figure at phase S7 is 211.
Merely as we had induced in portion 1 of this portfolio. the 2nd difference indicates that the general statement for the sequence will be a quadratic in the signifier: Sn=an2+bn+c Merely as we had done in portion 1. in order to obtain the general statement for the 6-stellar figure at phase Sn in footings of n. we can merely work out for it by a agency of the riddance method in order to acquire the values for the invariables ‘a’ ‘b’ and ‘c’ .













Equation # 1 – For phase 1: S1=an2+bn+c
1=12a+1b+c
1=a+b+c
Equation # 2 – For phase 2: S2=an2+bn+c
11=22a+2b+c
11=4a+2b+c
Equation # 3 – For phase 3: S3=an2+bn+c
31=32a+3b+c
31=9a+3b+c
First we will deduct Equation # 1 from Equation # 2 in order to extinguish ‘c’ :
11=4a+2b+c
1=a+b+c
The undermentioned equation ( we will name it Equation # 4 ) is obtained: 10=3a+b
The same will be done by deducting Equation # 2 from Equation # 3:
31=9a+3b+c
11=4a+2b+c














The undermentioned equation ( we will name it Equation # 5 ) is obtained: 20=5a+b
Now that we have eliminated the changeless ‘c’ from the equations. we will extinguish ‘b’ in order to work out for the value of ‘a’ by deducting Equation # 4 from Equation # 5:
20=5a+b
10=3a+b
The undermentioned equation is obtained: 10=2a
Solve for ‘a’ : a=5
Now we can replace the value of ‘a’ into Equation # 4 to work out for ‘b’ :
10=35+b
b=-5
Using the values of ‘a’ and ‘b’ and replacing them into Equation # 1 we can work out for ‘c’ :
1=5+-5+c
c=1










Now that we have defined the values for all 3 invariables. we can reason the general statement for the sequence by replacing the values into the primary quadratic equation:
General statement: Sn=5n2-5n+1 Factored signifier: Sn=5nn-1+1
Test the general statement:
Term 4: S4=544-1+1 Term 5: S5=555-1+1 The trials are valid. =61 =101
Therefore. the general statement that represents the 5-stellar figure at phase Tin: Sn=5nn-1+1



The first five representations for 4-stellar Numberss ( a star with 4 vertices ) are shown in the five phases 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 turning addition in the differences between each back-to-back set of Numberss of points. The difference between the figure of points in sets 2 and 1 is 8. the difference between the figure of points in sets 3 and 2 is 16. the difference between the figure of points in sets 4 and 3 is 24. and so on. The differences between each back-to-back brace of 4-stellar Numberss increase invariably by 8 each clip. This dictates that there is a changeless 2nd difference of 8 for this sequence. When the figure of points for each term in the sequence is observed. no forms can be clearly identified. However. by deducting 1 from each of the 4-stellar Numberss. the resulting Numberss are found to be factors of 8. Besides. by multiplying 8 by the amount of the n values from ( n-1 ) to 0. and adding the 1 we had subtracted earlier. the figure of points 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 form 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 form. an look for the 5-stellar figure at phase S7 can be obtained:
S7=86+5+4+3+2+1+1
=821+1
=169 The 4-stellar figure at phase S7 is 169.
Merely as we had induced in portion 1 of this portfolio. the 2nd difference indicates that the general statement for the sequence will be a quadratic in the signifier: Sn=an2+bn+c Merely as we had done in portion 1. in order to obtain the general statement for the 6-stellar figure at phase Sn in footings of n. we can merely work out for it by a agency of the riddance method in order to acquire the values for the invariables ‘a’ ‘b’ and ‘c’ .













Equation # 1 – For phase 1: S1=an2+bn+c
1=12a+1b+c
1=a+b+c
Equation # 2 – For phase 2: S2=an2+bn+c
9=22a+2b+c
9=4a+2b+c
Equation # 3 – For phase 3: S3=an2+bn+c
25=32a+3b+c
25=9a+3b+c
First we will deduct Equation # 1 from Equation # 2 in order to extinguish ‘c’ :
9=4a+2b+c
1=a+b+c
The undermentioned equation ( we will name it Equation # 4 ) is obtained: 8=3a+b
The same will be done by deducting Equation # 2 from Equation # 3:
25=9a+3b+c
9=4a+2b+c














The undermentioned equation ( we will name it Equation # 5 ) is obtained: 16=5a+b
Now that we have eliminated the changeless ‘c’ from the equations. we will extinguish ‘b’ in order to work out for the value of ‘a’ by deducting Equation # 4 from Equation # 5:
16=5a+b
8=3a+b
The undermentioned equation is obtained: 8=2a
Solve for ‘a’ : a=4
Now we can replace the value of ‘a’ into Equation # 4 to work out for ‘b’ :
8=34+b
b=-4







Using the values of ‘a’ and ‘b’ and replacing them into Equation # 1 we can work out for ‘c’ :
1=4+-4+c
c=1
Now that we have defined the values for all 3 invariables. we can reason the general statement for the sequence by replacing the values into the
primary quadratic equation:
General statement: Sn=4n2-5n+1 Factored signifier: Sn=4nn-1+1
Test the general statement:
Term 4: S4=444-1+1 Term 5: S5=455-1+1 The trials are valid. =49 =81
Therefore. the general statement that represents the 4-stellar figure at phase Tin: Sn=4nn-1+1







The first five representations for 7-stellar Numberss ( a star with 7 vertices ) are shown in the five phases 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 turning addition in the differences between each back-to-back set of Numberss of points. The difference between the figure of points in sets 2 and 1 is 14. the difference between the figure of points in sets 3 and 2 is 28. the difference between the figure of points in sets 4 and 3 is 42. and so on. The differences between each back-to-back brace of 7-stellar Numberss increase invariably by 14 each clip. This dictates that there is a changeless 2nd difference of 14 for this sequence. When the figure of points for each term in the sequence is observed. no forms can be clearly identified. However. by deducting 1 from each of the 7-stellar Numberss. the resulting Numberss are found to be factors of 14. Besides. by multiplying 14 by the amount of the n values from ( n-1 ) to 0. and adding the 1 we had subtracted earlier. the figure of points 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 form 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 form. an look for the 7-stellar figure at phase S7 can be obtained:
S7=146+5+4+3+2+1+1
=1421+1
=295 The 7-stellar figure at phase S7 is 295.












Merely as we had induced in portion 1 of this portfolio. the 2nd difference indicates that the general statement for the sequence will be a quadratic in the signifier: Sn=an2+bn+c Merely as we had done in portion 1. in order to obtain the general statement for the 6-stellar figure at phase Sn in footings of n. we can merely work out for it by a agency of the riddance method in order to acquire the values for the invariables ‘a’ ‘b’ and ‘c’ .

Equation # 1 – For phase 1: S1=an2+bn+c
1=12a+1b+c
1=a+b+c
Equation # 2 – For phase 2: S2=an2+bn+c
15=22a+2b+c
15=4a+2b+c
Equation # 3 – For phase 3: S3=an2+bn+c
43=32a+3b+c
43=9a+3b+c
First we will deduct Equation # 1 from Equation # 2 in order to extinguish ‘c’ :
15=4a+2b+c
1=a+b+c
The undermentioned equation ( we will name it Equation # 4 ) is obtained: 14=3a+b
The same will be done by deducting Equation # 2 from Equation # 3:
43=9a+3b+c
15=4a+2b+c














The undermentioned equation ( we will name it Equation # 5 ) is obtained: 28=5a+b
Now that we have eliminated the changeless ‘c’ from the equations. we will extinguish ‘b’ in order to work out for the value of ‘a’ by deducting Equation # 4 from Equation # 5:
28=5a+b
14=3a+b
The undermentioned equation is obtained: 14=2a
Solve for ‘a’ : a=7
Now we can replace the value of ‘a’ into Equation # 4 to work out for ‘b’ :
14=37+b
b=-7
Using the values of ‘a’ and ‘b’ and replacing them into Equation # 1 we can work out for ‘c’ :
1=7+-7+c
c=1










Now that we have defined the values for all 3 invariables. we can reason the general statement for the sequence by replacing the values into the primary quadratic equation:
General statement: Sn=7n2-7n+1 Factored signifier: Sn=7nn-1+1
Test the general statement:
Term 4: S4=744-1+1 Term 5: S5=755-1+1 The trials are valid. =85 =141
Therefore. the general statement that represents the 7-stellar figure at phase Tin: Sn=7nn-1+1



There has been a distinguishable relationship between the ‘p’ values and their corresponding general statements for the p-stellar Numberss. 5-Stellar: Sn=5nn-1+1
6-Stellar: Sn=6nn-1+1
7-Stellar: Sn=7nn-1+1
The coefficient in forepart of the ‘n’ in each general statement corresponds to the ‘p’ values. The remainder of the equation nevertheless remains unchanged as the figure of vertices ( ‘p’ value ) is changed. Therefore it can be induced that the general statement. in footings of ‘p’ and ‘n’ . that generates the sequence of p-stellar Numberss for any value of ‘p’ at phase Sn is: Sn=pnn-1+1


Now to prove 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 trials are valid. The general statement that generates the sequence of p-stellar figure for any value of ‘p’ at phase Sn is: Sn=pnn-1+1 However. there are limitations related to happening p-stellar Numberss. 1. The value of ‘p’ must be greater than 2

This is merely due to the fact that 2-stellar forms ( stars with 2 vertices ) or 1-stellar forms do non be. It is impossible to pull a star with 2 or fewer vertices. hence those values of P are inexistent and limitations must be indicated. 2. ‘p’ must be an component of a existent figure

Fanciful Numberss. such as -1 are non possible values for ‘p’ . They are inexistent and can non be created into leading forms. significance that there are no fanciful stellar Numberss that can be solved for. 3. The value of ‘p’ can non be rational

Leading forms are made of fit Numberss of vertices. Each vertex is counted as a individual whole figure. Rational Numberss of vertices can non be created due to it being impossible to hold a fraction of a vertex. There are besides limitations sing the term figure ( ‘n’ ) : 1. The term figure must be of positive whole number

Any sequence starts at 0. It is non possible to hold a sequence which begins in the negative part due to it being impossible to a negative term. It does non be. and will non bring forth important information by finding its inexistent term value. 2. ‘n’ must be an component of a existent figure

Like it is mentioned above. fanciful Numberss do non be. and can non be used as a term figure. 3. The value of ‘n’ can non be rational
For the same ground as to why there can non be rational ‘p’ values. term Numberss define a specific point in a sequence. Each term figure has a corresponding term value. However. rational term Numberss do non be. and have no term value. So the limitations on ‘p’ and ‘n’ are as follows:

3 ? p?? n?R p?Q 0 ? n?? n?R n?Q