• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
  1. 1
  2. 2
  3. 3
  4. 4
  5. 5
  6. 6
  7. 7
  8. 8
  9. 9
  10. 10
  11. 11
  12. 12
  13. 13
  14. 14

Math 20 Portfolio: Matrix

Extracts from this document...




By: Bob Bao

March 10, 2010



Triangular Pattern

Stellar Shape Pattern



In this portfolio, I will investigate a pattern of numbers that can be represented by a regular geometric arrangement of equally spaced points. The simplest examples of these are square numbers, 1, 4, 9, 16, which can be represented by squares of side 1, 2, 3 and 4. Other geometric shapes which can lead to special numbers are the triangular shapes and the stellar (star) shapes. Our calculations will be based on the sets of triangular and stellar diagrams that are already provided and those that will be constructed. The aim of this investigation is to examine and determine the general statement for geometric patterns that lead to special numbers as well as to demonstrate a good and clear understanding of patterns and the operations that can be done with them. At the end of the project, we should be able to generate expressions and recognise patterns of other various geometric arrangements.

Triangular Pattern:

In the following, I will examine how we can derive general statements of patterns within triangular shapes. To begin, we can use a sequence of diagrams to better illustrate the pattern:  image05.pngimage06.png








Firstly, looking at the pattern, we can see that the number of dots is the sum of a series of consecutive positive numbers. Considering this pattern within the triangular diagrams, we can use it to formulate the general formula. In the Math 10 pure curriculum, we were taught the sum of the series of consecutive 100 positive numbers. Accordingly, the sum of the first 100 positive numbers = 1 + 2 + 3... + 100 = (1+100) + (2+99) + ...+ (50+51) = (50)(101) = 5050.

...read more.




In each diagram, there are two copies of the triangular diagram, a black one and a white one. The original triangular diagrams from stage 1 to stage 8 now transformed into rectangles. It can be realized then that the number of dots of the original triangular diagram can be derived through the simple triangular area formula:


It is easy to see then that:

The 1st triangular number= (1 X 2) / 2 = 1

The 2nd triangular number= (2 X 3) / 2 = 3

The 3rd triangular number= (3 X 4) / 2 = 6

The 4th triangular number= (4 X 5) / 2 = 10

The 5th triangular number= (5 X 6) / 2 = 15

 The 6th triangular number= (6 X 7) / 2 = 21

 The 7th triangular number= (7 X 8) / 2 = 28

The 8th triangular number = (8 X 9) / 2 = 36

In general, then:  

tn = [n X (n + 1)] / 2

We can simplify the above general statement and arrive at:

tn = 0.5n2 + 0.5n

To have a correct general expression, it is essential that the statement be verified. In order to verify, we can use the already known value of 21 dots for the 6th stage. Therefore, simply substitute 6 for n into the derived general statement:

LS= 0.5(6)2 + 0.5(6)

RS= 21


For further insurance, simply substitute 3 for n into the general statement:

LS= 0.5(3)2 + 0.5(3)

RS= 6


By using advanced technology such as a graphing calculator, we can derive the general statement that represents the nth triangular number in terms of n much faster. The following table gives the number of evenly spaced dots in relation to the first seven stages of the pattern.


By plotting the first seven stages of the pattern on a graph, we can study the relation within the pattern much more effectively.

Graph settings are as follows:         X:[ 0, 10, 1]           Y:[0, 30, 1]


...read more.


Sn is limited to be greater and equal than 1, and is also a set of natural numbers. Since p is a parameter, it generally does not have a limitation. However, in the case of a geometric pattern, p > 0 since we cannot have a negative number of vertices in a geometric shape. As well, p cannot be 0 because the general formula would not have worked as the stellar number of every stage will become 1. To illustrate, when we substitute 0 for p, the general statement becomes 0n2 – 0n + 1 = 1. No matter what n is, the result will always be 1. Therefore, the derived general formula is invalid in a situation where p is 0.


Through analysing the patterns, the resourcefully use of a graphing calculator, the studying of the graphs and undertaking regression of the data, we can easily deduce the general statement of the two geometric patterns. The general statement of the triangular pattern is  tn = 0.5n2 + 0.5n where tn is the number of evenly spaced dots, and n is the nth stage. The general statement of the stellar pattern is Sn = pn2 – pn + 1 where Sn is the stellar number (number of dots), p is the number of vertices, and n is the nth stage. From the above investigation, we come to a conclusion that the general statements for any geometric shapes which lead to special numbers are quadratic functions with an upward concaving parabola that is only valid in the first quadrant of the graph where the Domain: image11.png, and Range: image27.png.

...read more.

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

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Maths essays

  1. Math IB HL math portfolio type II. Deduce the formula Sn = ...

    = (S10-S5) + (S20 - S15) 2 (60 - 80) = (80 - 60) + (S20 - 60) -40 = S20 - 40 S20 = 0 Therefore: S15 = 60 S20 = 0 6. For an arithmetic sequence, Sk = p and S2k = q.

  2. Extended Essay- Math

    �8M6�A�9� 3���$F�3/4�.Ø´Be�VD�s�W"(�/�.V�,!�}dhMz�'��b&�K�����"�&wE���:�K���\� X��4f������(tm)��I����"�R men��]��,*"F(y�$�^� ����� P˿��"ͺ��'7 N��*�-p���\��^��81/4>y�"�}-�,M�T���r<K�դ�(tm)x=Y���-j�x����At-��mXc�3/4x\&���'��t��3/4)�G�#C���L�^�`�� ��z`�%:k�Rz��c<(tm)� �rl�p'Q;���� �"��'�I$��(��sI�K��}y���fw�q�����x^�"�tKx���'"'b%�x�q�_"I���Ls�lw�ϳx1Ðg(��"��l��i� ��t�l �_|{Nb�E$E �4���d��Ç�� �;�v�%rQ1/4�{(tm)EG�|�E�t�,q�]���B��1��y2=���ߤ����V\"��9AXn�aX}�;���|�3�E-��;Z\A(r)]�V�� ��^�s �g� ��� ?�ϱ�k &�q�" ��.�4"X�����r�Z��NpR%�zV�o�(tm)...fâ¨ï¿½70�o�1^@B� ' Z��HL��H�F� 0z�'�� f���iJ<���|c8��[d�__�Í�p�/:(tm)�-z�+����cU �yݴ�R��ع<"VoY�)� K)�|�"�O���(�2I�MxÉB�ʪ'Iß²O�O����+)� �70�9D��k���3/4?'h!�4��-�(r)L��Da� �E�x;<4 �+@N��ɢ�-`�GB1/4Bb2��s*�m�Ë�:��"c HC,@`�e!5p3�s�m{� d�ZÖ-X�"Y�aNju�"�4���j?�-B�e� X�fB@y�1/2 -V"}Dd�14R�dOW@(Pt]H����1/2x�-x �vu<��\5�G��1/2�_$���MO�Ï>} 6��]�O�7���N"�D��jL�2+&�J)<"������.���|y%-Kx�Yk1/43<(c)�-Ô¿ �� �Nr=6�&Q�� � �"^A�)M�a~�I�3�N��(tm)�!�+(N��(tm)�l� ]�'y�Pf "j���8��IW��5\� 5- *�T� f��,�@�����<M l/�!�#A23/4�` �%@ �F�`8�r�8Õ°N'aR �� sÆ@i�(tm)��|���>^@��_...8��=�@[...ݦ�������m��"�S�i�'(c);� -�w�N���cK#-��"9�!4�o1�]t6�l��o

  1. Maths Internal Assessment -triangular and stellar numbers

    Question 6.2: Find an expression for the 6-stellar number at S7. Therefore, the expression for the 7-stellar number at stage S7 is (14 x 21)+1. Question 6.3: Find a general statement for the 7-stellar number at stage Sn in terms of n.

  2. Math Portfolio: trigonometry investigation (circle trig)

    a negative number is divided by a positive number resulting to a negative number. The value of y equals a positive number and the value of x equals to a negative number in quadrant 2. When the value of y is divided by the value of x, a positive number

  1. How many pieces? In this study, the maximum number of parts obtained by n ...

    = (2 + 4 + 8 + 16 + 31 + ...+k) + (k + 1) Substitute n = k case to get (1/24)k 4 - (1/12)k3 + (11/24)k2 + (7/12)k + 1 + (k + 1) In the form, Q = Z + Y + X + S, (Y + X + S)

  2. Ib math HL portfolio parabola investigation

    Forming my conjecture1: I can see that the value of D is related to the value of a. Accordingly, I can form a hypothesis that D = D(1) = = 1 --> TRUE D(2) = = 0.5 -->TRUE D(3) = = -->TRUE D(4)

  1. Stellar Numbers Investigation Portfolio.

    �2] x [2 x 2P + (n - 1 -1) 2P] Sn = S1 + [(n-1) �2] x [2 x 12 + (n - 2) 12] Sn = S1 + [(n-1) �2] x [24 + (n - 2) 12] Test the general statement for the 6-stellar numbers for multiple stages: General Statement: Sn = S1 + [(n-1)

  2. Stellar Numbers. In this study, we analyze geometrical shapes, which lead to special numbers. ...

    an outer star of a similar shape (with the same number of vertices). Thus, each figure at stage n consists of an n number of stars, placed inside one another similar to "Matryoshka dolls". Consider the features of the consecutive stars.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work