• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Stellar Numbers Investigation Portfolio.

Extracts from this document...

Introduction

Stellar Numbers

Triangular Numbers:

Diagram 1 : Triangular Numbers T1 = 1                T2 = 3                T3 = 6                T4 = 10                T5 = 15                T6 = 21

T7 = 28                T8 = 36

First Arithmetic Progression (AP) = T2 –T1

= 3 – 1

= 2

Difference from Arithmetic Progression (d) = (T3 – T2) – (T2 –T1)

= (6 – 3) – (3 – 1)

= 3 – 2

= 1

To determine the number of dots in one triangle there are two possible formulas which are possible. The first is that where Tn is equal to that of the first triangle, with the addition of the Sum of n -1. This formula can be simplified after substituting the Arithmetic Progression and difference into the Sum of n – 1. This gives the final formula in which Tn equals one (1) and n-1 divided by two (2) multiplied by four (4) and n minus two (2).

The other formula is that of when Tn is equal to n plus n-1 divided by two (2) multiplied by, two AP plus d of n take two (2). This formula can also be simplified and this gives the result of Tn equal to that of n divided by two, multiplied by that of n with the addition of one (1).

Tn = T1 + (Sumn – 1)

Tn = T1 + [(n-1) ÷2] x [2 x AP + (n – 1 -1) d]

Tn = 1 + [(n-1) ÷2] x [2 x 2 + (n – 2)]

Tn = 1 + [(n-1) ÷2] x [4 + (n – 2)]

Or

Tn = n + [(n-1) ÷2] x [2 x T1 + (n – 1 -1) d]

Tn = n + [(n-1) ÷2] x [2 x 1 + (n – 1 -1)1]

Tn = n + [(n-1) ÷2] x [2 + (n – 2)]

Tn = (2n + n2 –n) ÷ 2

Tn = (n + n2) ÷ 2

Tn = (n ÷ 2) (n + 1)

Table 1: Triangular Numbers

 Unit (n) Number of dots in triangle (Tn) (Tn+1) - (Tn) Progression 1 1 2 1 2 3 3 1 + 2 3 6 4 1 + 2 + 3 4 10 5 1 + 2 +3 +4 5 15 6 1 + 2 + 3 +4 +5 6 21 7 1 + 2 + 3 + 4 + 5 + 6 7 28 8 1 + 2 + 3 + 4 + 5 + 6 + 7 8 36 9 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8

Middle

The Arithmetic Progression for the 6-point stellar star is that of twelve (12), or two (2) multiplied by the number of points of the star in such case being that of six (6). Therefore AP is equal to 2p.

Find the expression for the 6-stellar number at stage S7:

Sn = S1 + [(n-1) ÷2] x [2 x 2P + (n – 1 -1) 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) 12]

S7 = 1 +  x [24 + 60]

S7 = 1 + 3 x 84

S7 = 1 + 252

S7 = 253

Find a general statement for the 6-stellar number at stage Sn in terms of n:

Sn = S1 + [(n-1) ÷2] x [2 x 2P + (n – 1 -1) 2P]

Sn = S1 + [(n-1) ÷2] x [2 x 12 + (n – 2) 12]

Sn = S1 + [(n-1) ÷2] x [24 + (n – 2) 12]

Test the general statement for the 6-stellar numbers for multiple stages:

General Statement: Sn = S1 + [(n-1) ÷2] x [24 + (n – 2) 12]

Stage S3:

S3 = S1 + [(3-1) ÷2] x [24 + (3 – 2) 12]

S3 = 1 + [2 ÷2] x [24 + (1) 12]

S3 = 1 +  x [24 + 12]

S3 = 1 + 

S3 = 37

Using the general statement it is possible to calculate that S3 of the 6- Stellar star has 37 dots in it. From the Diagram of the 6 point Stellar Star it is possible to show the formula is true for S3.

Stage S4:

S4 = S1 + [(4-1) ÷2] x [24 + (4 – 2) 12]

S4 = 1 + [(3) ÷2] x [24 + (2) 12]

S4 = 1 + [(3) ÷2] x [24 + 24]

S4 = 1 + [(3) ÷2] x 

S4 = 1 + 72

S4 = 73

The general statement makes it is possible to calculate that S4 of the 6- Stellar star has 73 dots in it. The Diagram of the 6 point Stellar Star it is possible to show the formula is true for S4.

Conclusion

Stage S3:

S3 = S1 + [(n-1) ÷2] x [28 + (n – 2) 14]

S3 = 1 + [(3-1) ÷2] x [28 + (3 – 2) 14]

S3 = 1 + [2 ÷2] x 

S3 = 43

Using the general statement it is possible to calculate that S3 of the 7- Stellar star has 43 dots. By continuing diagram 5 it is possible to support this outcome of produced from the general statement.

Limitations:

There are many faults with this general statement. Although it applies to many different numbered points of Stellar Stars, for it to be applied it is necessary to determine that the star has the same formation as that of the star which the general statement was created for.  In order to determine this it is necessary to determine that the arithmetic progression is equal to double that of the number of points of the star.

The number of points or P cannot be a negative number as this is not possible to form a negative star shape. The stellar shapes also only work when the number of point or vertices are equal to or greater than three (3). Thus it can be concluded that the formula only works for all possible stellar formation shapes.

This general statement also assumes that the first stellar unit is equal to that of one. Meaning that if the sequence of numbers started at any other numerical value the general formula may not apply. Therefore if your sequence did not include the first few layers it is possible the formula would not work as the S1.

 Kayla Jackson- Math Assessment Term 2- Standard Level- Mrs Moreno

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

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related International Baccalaureate Maths essays

1. ## Extended Essay- Math

�_j���@P@P@P@P@P@P@P@P@P@z�"���"�':E.����&JG%夶��(� f� �8� ;���^�s��\�{{��W�--""���h��qsl�...�`�[G-�VA�p"�;j( �� ��"8x����_��5��(��G_�2�M�������U��@�P@P@P@P@P@P@P@P@P@P@P � {��<Y�g �/�N��"]��� �&�s�y�k���P � � ( � ( � ( � ( � ( � ( � ( � ( � ( � ( � (�{��1/2�� -,������Mr�?�.�?�� "�"a����1/4�5��}�P � @P@P@P@P@P@P@P@P@P@P@�1/2������| �Ӧ�@�@��Bɰ���^~��3/4�(�� � ( � ( � ( � ( � ( � (

2. ## Maths Internal Assessment -triangular and stellar numbers

S1 S2 S3 S4 S5 S6 Layer 1 1 1 1 1 1 1 Layer 2 0 14 14 14 14 14 Layer 3 0 0 28 28 28 28 Layer 4 0 0 0 42 42 42 Layer 5 0 0 0 0 56 56 Layer 6 0 0

1. ## Stellar numbers

With this in mind, in order to account for this, what must happen is that the "n+1" must change into a "n-1" sign. This is necessary as by moving the graph to the left, the parabola will fit the data as demonstrated below: Figure 3: Graph of general statement of stellar shapes (6-vertices)

2. ## Math Portfolio: trigonometry investigation (circle trig)

When we put a random angle from quadrant 4, the range of 270<?<360, 336� in trial to verify the conjecture, the value of cos turn out to be positive while the values of sin and tan turn out to be negative.

1. ## Stellar Numbers Portfolio. The simplest example of these is square numbers, but over the ...

I decided to try for 5-stellar numbers, that the formula would be 5n2-5n+1. After plugging in numbers to this formula until up to S8 and seeing that all of them worked, the validity of this formula had been tested and proven.

2. ## Function Transformation Investigation

validity of the generalisations that have been made, we can try yet another function. First of all we identify the parent function of being of the form. Then, to make it be in the form We expand it: Now all we have to do is apply the rules that we

1. ## Stellar Numbers. Aim: To deduce the relationship found between the ...

of shapes so as to validate our hypothesis: 5 6 Now, let us put this information in a table: Term Dots 1 1 2 13 3 37 4 73 5 121 6 181 We can then find the relation between the values in the table by using the computer program

2. ## Lacsap triangle investigation.

Thus the graphing method complies with the equation found using the algebra; hence it can be concluded that the equation for the numerator is correct. Now, let us look at the trend for the denominator. The given triangle is symmetrical, meaning, the numbers on the left side have identical and • Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to 