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

Investigating a sequence of numbers

Extracts from this document...


Type 1: Investigating a sequence of Numbers

This is an investigation about series and sequences involving permutations. From a given series, I find the pattern of numbers that result from different values and use graphs to conjecture an expression from the series. By using mathematical induction and direct proof, I prove the general terms that I derived for the series.  

Part 1:

The sequence of numbers image00.pngis defined by

image01.png, image66.png, image75.png, …

image40.pngFrom the pattern of different values of n in image83.png above, I conclude thatimage92.png!

Part 2:


        If n=1

image100.png = image112.pngwere image02.png!



        If n=2

image29.pngwere image02.png!, image42.png!


image61.png+ (2 image64.png



        If n=3

image68.pngwere image02.png!, image42.png!,image69.png!


image71.png+ (2 image72.png



Part 3:

From Part 2, I know that:


To conjecture an expression of image39.png, I first organize the results that are derived in Part 2 to discover a pattern in the value of image39.pngas n increases.

Table 1.1:






















The same results of image78.png from Table 1.

...read more.


image87.png is true

  1. If image88.png is true, then


        If k=k+1, then


        Now, image91.pngimage93.png

                      = image94.png



                      =(k+2)!-1   image58.pngimage97.png (k+2)!

Thus image98.pngis true whenever image99.pngis true and image100.png is true.

image101.png is true for all n

Part 5:

image58.pngimage92.png!, I use this formula to show that image103.png is also true by simplifying image104.png to equal image105.png!

image105.png! = (n+1)!-n!

                   = (n+1)n!-n! image58.png (n+1)! = (n+1)n!

                   = n! (n+1-1)

                   = image105.png!


 Now, I use image103.png to device a direct proof for the expression of image81.pngthat I conjectured in Part 3.





From Part 3, I know thatimage110.png


When the first value of the first term is subtracted from the second value of the following term, 0 is derived so I cancel these terms. After I cancel the values to the most simplified manner, -1! and (n+1)! are left in the expression from which the following equation is derived:


image40.pngThe conjecture of image81.pngis proven

Part 6:

...read more.


. From the two graphs, I notice that (n+1)!(n+3) is exactly greater by 3 units to image31.png for all three points image40.png I conclude that:image39.png = (n+1)!(n+3) -3image38.png

Graph 2.1

Part 9:

The conjecture that I derived in Part 7 for image27.png can be proven through Mathematical Induction:

image31.png is: image41.png

image43.png + (2+2)!-2! + … + image44.png

  1. If n=1

         LHS: image45.png

             = (1+2)!-1!

             = 5

                     RHS: (1+1)!(1+3)image33.png= 5


image47.png is true

  1. If image48.png is true, then


image50.png + (2+2)!-2! + … + image51.png

        If k=k+1, then


        Now, image54.png



                = (k+1)!(k+2) image57.png

                = (k+2)!(k+4)image33.pngimage58.pngimage59.png

Thus image60.pngis true whenever image48.pngis true and image62.png is true.

image63.png is true for all n


Through this investigation, I have developed my knowledge about series and sequences involving permutations. I have learnt to use the patterns in a series to conjecture an expression for it and I had an opportunity to utilize my awareness of mathematical induction into proving the general term for the series. Most importantly, I have learnt to use technology related to series involving permutations. I enjoyed this investigation.


...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 IA type 2. In this task I will be investigating Probabilities and investigating ...

    When simplified the probability of deuce win for the player C becomes. Now I will use the generalized formulas above and technology in the form of Microsoft excel to find the probabilities of winning for values of c as 0.5, 0.55, 0.6, 0.7 and 0.9.

  2. Stellar Numbers. In this study, we analyze geometrical shapes, which lead to special numbers. ...

    Lets consider the diagram for the increased number of dots from Task 3: We can draw similar diagrams for stellar shapes with three or four vertices. These diagrams demonstrate the new dots added to the outer simple star at each consecutive stage (from S2 to S3 in this particular case).

  1. Stellar numbers

    and generated data, with data of the 7 stages of the stellar shapes and general statement which fits known data Based on the work above, the general statement for the 6-vertices star shapes are: Other Stellar Numbers (5-vertices, 7-vertices) In order to obtain a general statement other stellar shapes must

  2. Arithmetic Sequence Techniques

    1)d] an-1 = a1 + (n - 1 - 1)d] an+1 = a1 + (n + 1 - 1)d] an-1 + an+1 = 2a1 + (2n - 2)d = 2 [a1 + (n - 1)d] = 2an Therefore: 2an = an-1 + an+1 II.

  1. Stellar Numbers. In this task geometric shapes which lead to special numbers ...

    + 1 = 21 --> Using n=7: n2 + n + 1 = 36 (7)2 + (7) + 1 = 36 With proof that the formula works it is concluded that the general statement for this pattern is: n2 + n + 1 2.

  2. Stellar numbers. This internal assessment has been written to embrace one of the ...

    b=.5 c=0 Figure 2 Figure 2, to the left, shows what the TI-84’s display should look like after following the previous steps above.By substituting the appropriate values for a, b, and c, the following equation is formulated: y= 12x2+ 12x The general statement has to be in terms of n, not x¸ but beforehand, this equation can be further simplified.

  1. Stellar Numbers. In this folio task, we are going to determine difference geometric shapes, ...

    Curve is increasing. 6-STELLAR Now another stellar shape (star) with (6) p vertices, leading to (6) p-stellar numbers will be involved. Firstly, we are going to plot the stellar shape S5 and S6 by using the computer program ?Stellar. Jar?. S5 S6 After that, we can create another table for the stellar numbers, n 1 2 3 4 5 6 Sn 1 13 37 73 121 181 + 12(X)

  2. Math Portfolio Stellar Numbers. This task is an investigation of geometric numbers, the ...

    General term = Substitute n with 8 = = =36 The number of dots for 'n=8' is 36 which is identical to the value projected in the table of values. Square Numbers Figure 2: Table 2: Term number (n) Number of dots No.

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