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

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 the number of dots is 37 and in S4 the number of dots is 73.

The complete pattern in this 6-stellar is at follows:

0+1= 1

1+12=13

13+24=37

37+36=73

Therefore,

S5 = 73+ 48=121

S6 = 121 + 60= 181

Similarly to the triangular numbers, the 6-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 12 (as shown in red). Other way to write the pattern is in this way:

(1+0(12)), (1+1(12)), (1+3(12)) , (1+6(12)) , …

As soon as I wrote it this way I realized that there is clear relationship between the ...