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# Math Portfolio - Triangular Numbers

Extracts from this document...

Introduction

MOCK MATH PORTFOLIO

Triangular Numbers

1               3                          6                                  10

In a sequence of triangular numbers, the dots (as shown in the diagram above) represent a value of 1. In the sequence, the following triangle always has one more row of dots than the previous, and the first triangle has only one row of one dot. Each new last row in the next triangle always has one more dot than the last row in the previous triangle.

Since the first triangular number’s last row contains 1 dot, the next triangular number’s (the second triangular number’s) last row would contain 2 dots, and continue as such, and the number of any particular triangle in the sequence (the n value) would represent the number of dots in the last row of that triangle. Thus, in a continuation of the diagram, the fifth triangular number would be equivalent to the fourth triangle with an addition of a fifth row which contains five dots.

To complete the triangular number sequence with three more terms

Middle

= 2,  = 2×2×2 = 8

When  = 3,  = 3×3×3 =27

When  = 4,  = 4×4×4 = 64

When  = 5,  = 5×5×5 = 125

When  = 6,  = 6×6×6 = 216

When  = 7,  = 7×7×7 = 343

When  = 8,  = 8×8×8 = 512

When  = 9,  = 9×9×9 = 729

When  = 10,  = 10×10×10 = 1000

 n 1 2 3 4 5 6 7 8 9 10 1 8 27 64 125 216 343 512 729 1000

Let  be the sum of the first n cubic numbers.

Finding  for 1 ≤ n ≤ 10, where n is an integer can be calculated using technology.

Calculating  for 1 ≤ n ≤ 10 refers to finding the sum of all cubic numbers where n is greater than or equal to 1, and less than or equal to 10.

 n 1 2 3 4 5 6 7 8 9 10 1 9 36 100 225 441 784 1296 2025 3025

→ Calculated with the use of the GDC.

The general statement used to calculate  can be found as such:

 … …

Assume the function  exists such that:

=

=

=

…

=

=

In the function, the cubic number sequence would be the first difference of the terms of the function.

 … …

When finding, the sum can also be written as such:

= + + + … + +

= (+ ()+ ()+ … + ()+

(

= - + - + -+ … +- +-

=  -+ - + -+ … +- +-

= -

This is because when written in the form = (+ ()+ ()+ … + ()+ (, the first term in the first/previous bracket, cancels out the second term in the subsequent bracket, thus leaving only the second term of the first bracket, ...read more.

Conclusion

=

With no n value, the function cannot be used for. As n will be lowered by one degree, the appropriate function would have a polynomial with one more degree than = .

Let

=

=

+

+

+

+

=

When = -

=-

=

=                                 →

+                         →

+                 →

+                 →

+                 →

=

Since, the coefficient of must equal 1 so that coefficient) =:

Since only should remain, The remaining coefficients of n (, and) should equal 0. We can use  to solve for , , and

= + + + … + +

= -

=

=

=

=

=

=+

=+

=

=

=

=

=

 n 1 2 3 4 5 10 25 50 100 1 9 36 100 225 3025 105625 1.63e+06 2.55e+07

The validity of the general statement can be tested using technology.

The general statement is valid for all positive values of n (represented on the x-asis).

The scope/limitations of the general statement are as such:

The general statement works for all values of  n ≥ 0

The general statement was derived through the method of induction and the formation of conjectures.

The relationship between  and  can be seen as such:

=

=

=

= ,      =

This can be verified with the use of technology:

n

1

2

3

4

5

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

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