- Level: International Baccalaureate
- Subject: Maths
- Word count: 847
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
Middle
=1∙2∙3+…+5∙6∙7 = 210+210 = 420
...read more.
T6 =1∙2∙3+…+6∙7∙8 = 420+336 = 756
Tk =1∙2∙3+…+n(n+1)(n+2)
B) Tn = 
= 
= 
= 
= 
∴Tn = 
C) 1∙2∙3+2∙3∙4+3∙4∙5+…+n(n+1)(n+2) = 
i) n=1
1∙2∙3 = 
6 = 6
ii) n=k
1∙2∙3+2∙3∙4+3∙4∙5+…+k(k+1)(k+2) = 
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) = 
+ (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) = 
1∙2∙3+2∙3∙4+3∙4∙5+…+k(k+1)(k+2)+(k+1)(k+2)(k+3) = 
iii) n=k+1
1∙2∙3+2∙3∙4+3∙4∙5+…+(k+1)(k+2)(k+3) = 
D) 13+23+33+43+…+n3
= 
= 
= 
= 
=
= 
∴ =
Question 4
A)
U1 =1∙2∙3∙4 = 24
Conclusion
+1)(k+2)(k+3)+ (k+1)(k+2)(k+3)(k+4) = 
...read more.
iii) n=k+1
1∙2∙3∙4+2∙3∙4∙5+3∙4∙5∙6+…+(k+1)(k+2)(k+3)(k+4) = 
D) 14+24+34+44+…+n4
= 
= 
∴ 
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 
and put those coefficients to
(n+1)k+1–nk+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
∴ = 1 k+2 k+3 k+4 k+…+n k =
{(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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This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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