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Math Portfolio - Triangular Numbers

Extracts from this document...

Introduction

MOCK MATH PORTFOLIO

Triangular Numbers image00.jpg

                           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

...read more.

Middle

 = 2, image47.pngimage47.png = 2×2×2 = 8

        When image44.pngimage44.png = 3, image48.pngimage48.png = 3×3×3 =27

When image44.pngimage44.png = 4, image49.pngimage49.png = 4×4×4 = 64

When image44.pngimage44.png = 5, image51.pngimage51.png = 5×5×5 = 125

When image44.pngimage44.png = 6, image52.pngimage52.png = 6×6×6 = 216

When image44.pngimage44.png = 7, image54.pngimage54.png = 7×7×7 = 343

When image44.pngimage44.png = 8, image55.pngimage55.png = 8×8×8 = 512

When image44.pngimage44.png = 9, image56.pngimage56.png = 9×9×9 = 729

When image44.pngimage44.png = 10, image58.pngimage58.png = 10×10×10 = 1000

n

1

2

3

4

5

6

7

8

9

10

image59.png

1

8

27

64

125

216

343

512

729

1000

Let image60.pngimage60.png be the sum of the first n cubic numbers.

Finding image61.pngimage61.png for 1 ≤ n ≤ 10, where n is an integer can be calculated using technology.

Calculating image62.pngimage62.png 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

image63.png

1

9

36

100

225

441

784

1296

2025

3025

→ Calculated with the use of the GDC.

The general statement used to calculate image60.pngimage60.png can be found as such:

image64.png

image65.png

image66.png

image67.png

image68.png

image69.png

image71.png

image72.png

image73.png

image74.png

image75.png

image76.png

image64.png

image30.png

image77.png

image32.png

image78.png

image34.png

image35.png

image79.png

image80.png

image81.png

image82.png

Assume the function image83.pngimage83.png exists such that:

image45.pngimage45.png = image85.pngimage85.png

image47.pngimage47.png = image86.pngimage86.png

image48.pngimage48.png = image87.pngimage87.png

           …

image88.pngimage88.png = image89.pngimage89.png

image90.pngimage90.png = image91.pngimage91.png

        In the function, the cubic number sequence would be the first difference of the terms of the functionimage92.pngimage92.png.

image94.png

image95.png

image96.png

image97.png

image98.png

image99.png

image100.png

image101.png

image94.png

image103.png

image64.png

image65.png

image66.png

image67.png

image68.png

image69.png

image76.png

image64.png

When findingimage104.pngimage104.png, the sum can also be written as such:

image106.pngimage106.png = image45.pngimage45.png+ image47.pngimage47.png+ image48.pngimage48.png+ … +image107.pngimage107.png +image108.pngimage108.png

= (image109.pngimage109.png+ (image110.pngimage110.png)+ (image111.pngimage111.png)+ … + (image112.pngimage112.png)+

   (image114.pngimage114.png

                = image115.pngimage115.png- image116.pngimage116.png+ image117.pngimage117.png- image115.pngimage115.png+ image118.pngimage118.png-image119.pngimage119.png+ … +image120.pngimage120.png- image121.pngimage121.png+image122.pngimage122.png-image124.pngimage124.png

= image125.pngimage125.png -image116.pngimage116.png+ image126.pngimage126.png- image127.pngimage127.png+ image128.pngimage128.png-image129.pngimage129.png+ … +image130.pngimage130.png- image131.pngimage131.png+image132.pngimage132.png-image134.pngimage134.png

= image135.pngimage135.png-image136.pngimage136.png

This is because when written in the formimage104.pngimage104.png = (image137.pngimage137.png+ (image138.pngimage138.png)+ (image139.pngimage139.png)+ … + (image140.pngimage140.png)+ (image141.pngimage141.png, 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, image142.png...read more.

Conclusion

   =image156.pngimage156.png

        With no n value, the function cannot be used forimage92.pngimage92.png. As n will be lowered by one degree, the appropriate function would have a polynomial with one more degree than image41.pngimage41.png= image42.pngimage42.png.

Let image158.pngimage158.png

image159.pngimage159.png = image160.pngimage160.png

                       = image161.pngimage161.png

                         +image162.pngimage162.png

                        + image163.pngimage163.png

                        + image164.pngimage164.png

                        + image165.pngimage165.png

      = image166.pngimage166.png

        When image108.pngimage108.png= image132.pngimage132.png-image124.pngimage124.png

image167.pngimage167.png =image168.pngimage168.png-image124.pngimage124.png

   =image169.pngimage169.png

                   = image170.pngimage170.png                                →image171.pngimage171.png

                        + image172.pngimage172.png                        → image173.pngimage173.png

                        + image174.pngimage174.png                →image175.pngimage175.png

                        + image176.pngimage176.png                →image177.pngimage177.png

                        + image178.pngimage178.png                →    image179.pngimage179.png

                   = image180.pngimage180.png

Sinceimage181.pngimage181.png, the coefficient of image182.pngimage182.pngmust equal 1 so that image183.pngimage183.pngcoefficient) =image185.pngimage185.png:

image186.png

image187.pngimage187.png

        Since only image42.pngimage42.pngshould remain, The remaining coefficients of n (image188.pngimage188.pngimage189.pngimage189.png, andimage190.pngimage190.png) should equal 0. We can use image191.pngimage191.png to solve for image192.pngimage192.png, image193.pngimage193.png, and image195.pngimage195.png

image196.pngimage196.png

image197.pngimage197.png

image198.pngimage198.png

image199.pngimage199.png

image200.pngimage200.png

image201.pngimage201.png

image202.pngimage202.png

image203.pngimage203.png

image204.pngimage204.png

image205.pngimage205.png

image206.pngimage206.png

image207.pngimage207.png

image208.pngimage208.png

image209.pngimage209.png

image210.pngimage210.png

image211.pngimage211.png

image212.pngimage212.png = image45.pngimage45.png+ image47.pngimage47.png+ image48.pngimage48.png+ … +image107.pngimage107.png +image108.pngimage108.png

             = image135.pngimage135.png-image136.pngimage136.png

             = image214.pngimage214.png

             =image215.pngimage215.png

             =image216.pngimage216.png

             =image217.pngimage217.png

             =image218.pngimage218.png

             =image219.pngimage219.png+ image220.pngimage220.png

             =image221.pngimage221.png+ image220.pngimage220.png

=image222.pngimage222.png

             =image224.pngimage224.png

             =image224.pngimage224.png

             =image225.pngimage225.png

     =image226.pngimage226.png

n

1

2

3

4

5

10

25

50

100

image227.png

1

9

36

100

225

3025

105625

1.63E6

2.55E7

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).

image228.png

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 image229.pngimage229.png and image60.pngimage60.png can be seen as such:

image230.pngimage230.pngimage231.pngimage231.png

image232.pngimage232.png = image233.pngimage233.png

image234.pngimage234.png

image235.pngimage235.png = image236.pngimage236.png

image235.pngimage235.png = image231.pngimage231.png

image229.pngimage229.png = image237.pngimage237.png,     image60.pngimage60.png = image238.pngimage238.png

This can be verified with the use of technology:

n

1

2

3

4

5

...read more.

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

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