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IB Math HL Portfolio Type I Series and Induction

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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

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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

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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.

        -  -

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