• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

IB Math HL Portfolio Type I Series and Induction

Extracts from this document...

Introduction

Mathematics Higher Level Portfolio Type I Series and Induction

Acknowledgement

Question sheet should directly be given from your IB Mathematics HL teachers. This is due to the fact that IB students are not allowed to hold any question paper; every candidate must finish this coursework and return the question sheet in five days. Therefore I was not able to include any questions in this portfolio.

Introduction

This Investigation of the Series and Induction Portfolio for Math HL brings out that the sum of terms of series following a certain pattern can be predicted as expressions by studying these patterns. With the resourceful use of a calculator, studying the graphs and

...read more.

Middle

=1∙2∙3++5∙6∙7 = 210+210 = 420

T6 =1∙2∙3++6∙7∙8 = 420+336 = 756

Tk =1∙2∙3++n(n+1)(n+2)

B) Tn = image18.png

= image19.png

= image20.png

= image21.png

= image22.png

Tn = image23.png

C) 1∙2∙3+2∙3∙4+3∙4∙5++n(n+1)(n+2) =  image23.pngimage24.png

i) n=1

1∙2∙3 = image25.png

      6 = 6

ii) n=k

1∙2∙3+2∙3∙4+3∙4∙5++k(k+1)(k+2) =  image26.pngimage02.png

Add (k+1)(k+2)(k+3) both side

1∙2∙3+2∙3∙4+3∙4∙5++k(k+1)(k+2)+(k+1)(k+2)(k+3) = image28.png+ (k+1)(k+2)(k+3)

1∙2∙3+2∙3∙4+3∙4∙5++k(k+1)(k+2)+(k+1)(k+2)(k+3) = image29.png

1∙2∙3+2∙3∙4+3∙4∙5++k(k+1)(k+2)+(k+1)(k+2)(k+3) = image30.png

image06.png

iii) n=k+1

1∙2∙3+2∙3∙4+3∙4∙5++(k+1)(k+2)(k+3) = image30.pngimage07.png

D) 13+23+33+43++n3

image23.png= image31.png

image23.png= image32.png

image33.png= image34.png

          =  image35.png

          =  image35.png

          =  image36.png

image33.png=

Question 4

A)

U1 =1∙2∙3∙4 = 24

...read more.

Conclusion

+1)(k+2)(k+3)+ (k+1)(k+2)(k+3)(k+4)   = image51.pngimage52.png

iii) n=k+1

1∙2∙3∙4+2∙3∙4∙5+3∙4∙5∙6++(k+1)(k+2)(k+3)(k+4)   = image51.pngimage53.png

D) 14+24+34+44++n4

image45.png= image54.png

image45.png= image56.png

image57.pngimage58.pngimage59.png

image60.png

image61.pngimage62.png


Question 5

Use of the Pascal’s triangle

1

1  2  1

1  3  3  1

1  4  6  4  1

1  5  10 10 5  1

1  6  15  20 15  6  1

1  7  21  35  35 21  7  1

1  8 28 56  70  56  28  8  1

…… And so on

By using Pascal’s triangle, we can get the results with binomial theorem.

We could use this triangle to get image63.png and put those coefficients to image63.png

(n+1)k+1nk+1= ank+bnk-1+cnk-2+1

n=1

2k+1–1k+1= a∙1k+b∙1k-1+c∙1k-2+1

n=2

3k+1–2k+1= a∙2k+b∙2k-1+c∙2k-2+1

n=m

(m +1)k+1 m k+1= ank+bnk-1+cnk-2+1

                           = a(1 k +2 k +3 k ++ m k ) + b(1 k-1 +2 k-1 +3 k-1 ++ m k-1)+ …+ m

image63.png = 1 k+2 k+3 k+4 k++n k =image64.png{(m+1) k+1–1–b(1 k-1 +2 k-1 +3 k-1 ++ m k-1)+–m}

Those a, b, c, d represents the coefficients. We can simply get the coefficient from the Pascal’s triangle.

        -  -

...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 = ...

    + (a2 + a n-1) + (a3 + a n-2) = 15 3 (a1 + an) = 15 a1 + an = a2 + an-1 = ... = an + a1 (a1 + an) = 5 Sn = n (a1 +an)

  2. Extended Essay- Math

    ��#l��o���� ͳy"��mß'��Ì�*...F�'�- �e-jADN�2l���"��gtñm�O�Ka_�=M��n�|�&q�i6�)��(tm)$��1��ĴaÙ�m0p% � �"h5a"G���...�bz�U>�95�D��s�qÌ¢!�-� 0 T...Bbe,��ÎJ"� ]!g"�b1/4t����3/4�qk��Y��9�3/4�v�r�]�4tҦ��&��[�9N�sR���|� h(tm)�d����)��c�_�' }M1/4"j ��<Z@��G�(tm)1/4*�Ï��Í� $��j� �*(tm)�9�K�-1��˦up���;% ��bt�'��3/4��[��u��)J�It"p"�-�hn�t�$�� �m?c ?�...{`"b/�m���� Ò8Flk�&fSh2stj[v�;�k��c�hW��*��O0d�- ��0e�-�N ~'zI���_ß�>Ds�Ù�� �=-_d��0�E1^�*�J�aXE+(tm)V�ng��R��Lq�C�"(r).z}�o6j��DNLk� �(tm)�(r)�4"�qXZÛ´4��F�&�U���Q���Q�...pU��Ö� )i�>o;_>4:lo�u0E>�T��Y:��1/2��C��ÞW(tm)1/4ZR#Pi��]LH_!~�-(tm)�� �-M�-� |�|�(c)...�� �= 'Ûrm�p�P���=�����?��1/2�o(��-�?7�*L�n6n(tm)v�߳h�.Hb;... '7}>'"�"_�q��nÚ窴=�c���]-�...4�f� \�5�H ��>�3/4�d��*I?���H |4SA������ ...�W�6-��8�[�bf -� �Ó���6�BHÂA�m(7�ӣ...?�-=�1/2�} |�_)�2L(c)ES_1l�I��}�f��#i�b��Ê��"�.m�ǰ�Ò#/Xi"!�L��� �J:���"�Q��i� S)kg\���� k� �qظ���"��O?A_���j"= .��~2...�H"p��i��o��'^'@��$�Uo �z�"RÕb9Ç°+"�t�=��b��$"ʥ�m=��(c)M��ԥ`������#�#Rz_ ��br�1/41/2 ���R@C(tm)��%tÞ¸1IV�`�z=�I�z- a���15TÐ��(tm)� A� 0~�[M m)�Υ�o�Y-�3/4 ˴�@Ar]�_-V�k"[~G'�e`ÄZ�"e� �?�-oa1/2...��a0��T�wm� E�+...�kK�ؿ%���'�͵K=�R...σ ^

  1. Math Portfolio Type II Gold Medal heights

    function and the system of equation is limited to two unknown only. See next page for the acquisition of b and k. One should not at this point that these points were not randomly chosen. They both represent anomalies and therefore do not fit the general pattern, however their effect

  2. Ib math HL portfolio parabola investigation

    (see Graph 1) Graph 1: Value of a=1 4. Then I found out the values of (X2 - X1) and (X4 - X3) and called them SL and SR respectively. 5. I finally calculated the value of D = |SL - SR| These results are illustrated in table 1.

  1. Tarea Tipo 1 (nm)

    enumerado es ; usando la formula y sustituyendo obtenemos los siguientes resultados: = = = 3 paralelogramos Ahora probemos con los siguientes ejercicios con los cuales sus resultados correspondientes ya fueron comprobados anteriormente con la f�rmula de conjuntos: a.

  2. IB Mathematics Portfolio - Modeling the amount of a drug in the bloodstream

    drug in the bloodstream would continue to decrease until the amount is too small to measure. The model is an exponential decay which means the line will never touches zero, but in this case, I can assume that the drug would eventually be completely gone from the body even though the model shows that the amount would never reach zero.

  1. MAths HL portfolio Type 1 - Koch snowflake

    To investigate what happens at stage four, we will have to apply all of the above equations and substitute n as 4. we will have to substitute in these for Stage (n) 4 This is how one side of the Koch snowflake would look at stage 4.

  2. Shadow Functions Maths IB HL Portfolio

    If we want to generalise an expression for the shadow function, we need to reconsider the definition of the shadow function in itself. It has the same vertex, but opposite concavity. Thus, it is reflected by the line passing through the vertex and horizontal to the x-axis: .

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