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# IB math portfolio - Triangular and Stellar numbers

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Introduction

Triangular numbers

The following  diagram shows a trinagular patterns of evenly spaced dots.The  number of dots in each diagram are examples of Triangular numbers :

In order to show each diagram I bring the name in terms of (T1,T2,T3..)  with number of dots in each first row of diagram. I notice that there is always one more row than in previous diagram , that row has one more dot than the previous one.With knowing that the difference between each diagram is the same, I assume that the diagrams are parts of the seguence. T1                  T2                 T3                        T4                                 T5

In that form, I can determine that next diagrams can be completed by adding  an extra row with the same number of dots as the term number to the prevous one . For example, T7 will have the same amount of dots as T6 plus Term number(15 +6=21).Therefore, to the next three diagrams I will give name in order with amount of dots in extra row : T6, T7, T8. T7                                         T8                                                T9

Middle

Stellar numbers

Stellar shapes – are the shapes with p vertices leading to p stellar numbers. The first four representations for the star with six verticers are shown in the four stages below( S1- S 4).The 6 stellar number at each stage is total number of dots in the diagram: S 1                     S 2                                S 3                                         S4

To find out the number of dots in each stage up to S6 I’m going to organize the data to figure out the difference between the diagrams:

 Term number S 1 S 2 S 3 S4 Number of dots 1 13 37 73 Number of layers 1 2 3 4

Unfortunately I can’t see the clear pattern from my first table , so I

Conclusion

3

S4

S5

S6

Number of dots

1

1+12

1+12+24

1+12+24+36

1+12+24+36

+48

1+12+24+36

+48+60

Therefore, I see that formula will work for S7 but It won’t work for another terms, because you always have to know the previous one.

I assume that I can use the same formula that I used with the triangular numbers but I need always add 1, because of the first term.

Using previous information I have Sn =  n(2a+d(n-1))/2  +1

I will use 12 as the first term(a) which is the same as S2 ( if you add 1), so that’s why n is going to be one less than 7 which is 6.

So I can substitute it using calculator : S7 = 6(2*12 +(7-1)12))/2 +1 = 253.

So now I have S7 as 1+12+24+36+48+60+72 = 253, which proves the pattern and appropriate for it.

Because the  value of a is S2  instead of the first term , all formula will be equal to S n+1:

Using all the knowledge I have

Sn+1   = n(2*12+12(n-1))/2  +1

Using algebra I simplified it into a formula :

S n+1 = 6n(n+1)+1

Therefore I think we need to represent n as n-1, because this equation is equal to S n+1 , so now I substitute and get this:

Sn+1 =6(n-1)( n-1+1)+1=6n2-6n+1.

This general statement can be applied for 6 stellar numbers

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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Since some of the values of Sn have already been established this makes it possible to work out the general formula. The first step is to substitute the established values into the three quadratic equations, as shown below: Sn = an2 + bn + c Therefore: quadratic When n = 1, Sn = 1 1 = a(1)2 + b(1)

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From the Diagram of the 5 point Stellar Star it is possible to show the formula is true for S2. Stage S3: S3 = S1 + [(n-1) �2] x [2 x 10 + (n - 2) 10] S3 = 1 + [(3-1)

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0.883022 1 -140 0.413175911 0.586824 1 -120 0.75 0.25 1 -100 0.96984631 0.030154 1 -90 1 3.75E-33 1 -80 0.96984631 0.030154 1 -60 0.75 0.25 1 -40 0.413175911 0.586824 1 -20 0.116977778 0.883022 1 0 0 1 1 20 0.116977778 0.883022 1 40 0.413175911 0.586824 1 60 0.75 0.25 1 • Over 160,000 pieces
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