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# Stellar Numbers Portfolio. In this task I will consider geometric shapes, which lead to special numbers

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Introduction

Vladislav Tajc                  Maths SL type 1 Portfolio: Stellar Numbers                             1/28/12

The English School

IB Mathematics SL

Math Portfolio (Type1)

Stellar Numbers

12th of September 2011

Bogotá, Colombia

Stellar Numbers

Aim: In this task I will consider geometric shapes, which lead to special numbers

Complete the triangular sequence with more than 3 terms.  Following the triangle sequence:

0+1=1

1+2=3

3+3=6

6+4=10

10+5=15

Therefore,

15+6=21

21+7=28

28+8=36

After looking at the sequence I could realize that these triangular numbers are simply the sum of numbers from 1 to the term number. As the pattern continues the adding number increases arithmetically.

Example: Triangular number 4 is 10 so, 6+4=10; or the Triangular number 7 is 28 so, 21+7= 28

Hence, the general statement that represents the  triangular number in terms of  is the equation:  , Were  is the triangular number we want to find

For example, if we want to find the 10th triangular number we replace  with 10  =  = 55

This means that the 10th term will have 55 dots in its triangular shape.

Find the number of dots (the stellar number) in each stage up to S6 At S1 the number of dots is 1, at S2 the number of dots is 13, at S3

Middle To find the general statement we must look back to the Triangular number equation of  , at this point we need to modify the equation so that it is suitable for the 6-stellar. So, in this case  =12n because of the multiples of 12, the (  to  because each of the numbers multiplied by 12 is one less than the  of the term, and add 1 because the addition of terms starts initially from 1. Following all the rules the general statement should look like this: Repeat the steps above for other values of  A) Here is the number of dots for the sequence of 5-stellar number

S1 = 0+1 = 1

S2 = 1+10= 11

S3 = 11+20= 31

S4 = 31+30= 61

Therefore,

S5 = 61+40= 101

S6 = 101+50= 151

Similarly to the triangular numbers, the 5-stellar sequences uses the same method but this time the number of dots can be found by preceding term added to by the multiples of 10 (as shown in red). Other way to write the pattern is in this way:

(1+0(10)), (1+1(10)), (1+3(10)) , (1+6(10)) , …

As soon as I wrote it this way I realized that there is clear relationship between the Triangular numbers and the way the 5-stellar numbers  term.

Conclusion

p and n is:

Sn = 1+ (  Test the validity of the general statement

Since we now have a general statement we can ensure that it works on larger numbers

For example: if we have  and we want to know S8, then  . Using our general statement we can get a quicker and easier answer:

S8 =  S8 = 1+ 840(28)

S8 = 1+ 23520

S8  = 23521                 (At S8 the 420-stellar will have this amount of points)

Explain how you arrived at the general statement

In order to arrive at the general statement I did first take the general statement of the triangular numbers on task 1 and then adjusted as a arithmetical sequence so that task 2 could be done. From the general statement of the 6-stellar number, I could now adjust the equation in order to get different values of p and its sequences. After doing the examples of 6,7 and 8 stellar numbers, I could therefore, understand the behaviour of the sequences by considering the relation of p and n. Following this relation, I was able to put either p or n in the general statement to by then take any data, no matter if it is a small or big value, as well as the geometrical shape of the number.

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

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means that the dots of two triangles is equal to one square. Then, as there are "n" quantities of "n+1" the equation becomes "n(n+1)". Then the whole equation is divided by 2 to make the equation representative for the triangle rather than the square.

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�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)

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171 243 Question 4 Find an expression for the 6-stellar number at stage S7 The expression which I used for the 6-stellar number at stage S7 is carried down from the . previous information determined that each term is 12n more than the previous term, in which n is equal to the term number of the previous term. • Over 160,000 pieces
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