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IB math portfolio - Triangular and Stellar numbers

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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

...read more.


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:image02.png

       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


Number of dots





Number of layers





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

...read more.






Number of dots









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

...read more.

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

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