Ib Math Sl 1

Only available on StudyMode
  • Topic: Triangular number, Polygonal number, Square number
  • Pages : 8 (1913 words )
  • Download(s) : 888
  • Published : March 13, 2012
Open Document
Text Preview
The task of this assessment is to consider and investigate geometric shapes that lead to stellar numbers. One of the simplest forms of this would be of square numbers, terms 1, 2, 3, 4, and 5 in terms of geometric shapes lead to the special numbers of 1, 4, 9, 16, and 25. To find these outcomes, simply all that is needed to be done is to square the nth term of the sequence. For example, 12=1, 22=4, 32=9, 42=16, and 52=25, therefore the resulting formula for square numbers would be n2=Tn where T=total amount of n. For this investigation though, the geometric shapes of equilateral triangles are studied. Triangular numbers are the amount of dots that evenly fit inside the triangle. After solving the sequence of the triangular numbers and finding a formula to solve for all equilateral triangles, the knowledge attained from triangular numbers will then be used to find and solve the correlation between triangular numbers and stellar (star) numbers.

Here is a sample of the first 5 terms in the triangular numbers sequence: The row of numbers above the triangles is the number of each term or n. The rows of numbers below the triangles are triangular numbers of each term or n. `n1n2n3n4n5

Here is a sample of the first 4 stages of p-stellar shapes that are being investigated. P indicates the amount of vertices in the stellar shape. In this case the stellar shapes being investigated are 6-stellar, however after deriving a formula, the stellar number of any p vertices may be found.

Sn is the different stellar shape terms, and the numbers aligned in a row below S1-S4 are the p-stellar numbers, in this case the 6-stellar numbers.


Triangular Numbers:

The 2 triangles on the right are n2 and n3
From n1 to n2, 2 dots are added making a 2nd row of dots, and from n2 to n3, 3 dots are added making a 3rd row of dots. This is repeated for every term in the triangular sequence; adding another row of dots, and the row added equals the term.

For example n2 has 3 dots total, and n3 has 6 dots total, in the transition from n2 to n3 another row of dots is added onto n2 extending the triangle. In this case n=3, there are 3 rows of dots in the triangle, and the row that has just been added has 3 dots in that row. Also when n=4, there are 4 rows of dots, and the last row has 4 dots in that row. Clearly there is a correlation between the nth triangular number and the amount of dots inside the triangle. Every term is equal to the sum of all “n’s” that precede it and itself. Mathematical procedures:



In the following process consider T to be the total number of dots in a particular triangle. The previous process can now be condensed into a formula. Tn=Tn-1+n T3=T3-1+3
Although the triangular numbers sequence has now been formed into a distinct pattern and formula, this formula is recursive and a non-recursive formula must be achieved. In this scenario non-recursive can be defined as a formula that does not rely on any previous knowledge of the total amount of dots in any given triangle.

Furthermore it is seen that Tn-1+Tn=the square number of n. For example n3=6, n4=10, 6+10=16 which is 42. By adding n to 16 the outcome is 20. 20 is then divided by 2 and the answer is 10 which is T4. N2=Tn-1+TnN=4Tn=10Tn-1=610+6=1616+4=2020/2=10

(n2+n)/2= Tn
This process results in the arithmetic sequence Tn=(n2+n)/2. To clearly understand the process, consider the example calculations on the following page.

Formula: Tn=(n2+n)/2

T2= (22+2)/2
tracking img