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Stellar Number This task purpose is to find a sequence of arithmatic or geometric series that are determine by a special geomatric shape , which in this case a stellar . To find the special pattern in the series I tried the task A first , which in this case a triangle made by circular shape which have a pattern of 1,3,6,10,15 and to continue the U N with three more terms the U N are 21,28,36. 1 2 3 4 5 6 7 8 Table 1.1 N U N Pattern D 1 1 1 1 2 3 1+2 2 3 6 1+2+3 3 4 10 1+2+3+4 4 5 15 1+2+3+4+5 5 6 21 1+2+3+4+5+6 6 7 28 1+2+3+4+5+6+7 7 8 36 1+2+3+4+5+6+7+8 8

Stellar Number( Math IA)

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Page 1: Stellar Number( Math IA)

Stellar Number

This task purpose is to find a sequence of arithmatic or geometric series that are determine by a special geomatric shape , which in this case a stellar . To find the special pattern in the series I tried the task A first , which in this case a triangle made by circular shape which have a pattern of 1,3,6,10,15 and to continue the UN with three more terms the UN are 21,28,36.

1 2 3 4 5 6 7 8

Table 1.1

N UN Pattern D

1 1 1 1

2 3 1+2 2

3 6 1+2+3 3

4 10 1+2+3+4 4

5 15 1+2+3+4+5 5

6 21 1+2+3+4+5+6 6

7 28 1+2+3+4+5+6+7 7

8 36 1+2+3+4+5+6+7+8 8

From the pattern above we know that the N are equal to the D which in this case are increasing in number by 1 , so we can say that “D = N” so the formula are Un=a+ (n−1 )n

Page 2: Stellar Number( Math IA)

If we apply that to the existing table it would look like this :

N UN Un=a+ (n−1 )n D

1 1 Un=1+(1−1 )1 1

2 3 Un=1+(2−1 )2 2

3 7 Un=1+(3−1 )3 3

4 13 Un=1+(4−1 )4 4

5 21 Un=1+(5−1 )5 5

6 31 Un=1+(6−1 )6 6

7 43 Un=1+(7−1 )7 7

8 57 Un=1+(8−1 )8 8

As we can see although the pattern of the first and second series are correct ,the rest are not so it can’t be applied after that