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Investigating a Sequence of Numbers.

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Introduction

Mathematics HL Portfolio: Investigation a Sequence of Numbers        2007/5/4        

Mathematics HL Portfolio Assignment

Investigating a Sequence of Numbers [Type 1]

        In this Mathematics Portfolio, I am going to investigate a sequence of numbers by mathematical methods which I have learnt in the I.B. Mathematics HL course. Throughout the investigation, I will include all my workings in order to let examiners know exactly how I come up with the answers.

        A sequence is a set of numbers with a definite order. A series is a sum of a sequence. The sequence of numbers {an}n =1 is:

1 x 1!, 2 x 2!, 3 x 3!, …

The two signs outside the bracket of an represent the range of the sequence. The bottom one is where the sequence begins and the one above is where it should end. Since it is stated the sequence starts from n = 1, therefore the first term, a1 = 1 x 1!, the second term, a2 = 2 x 2! and the third term, a3 = 3 x 3!……

The ! sign after the numbers is called a factorial notation. The notation basically means the product of all the numbers from 1 to the number with the notation. For example:

                                3! = 1 x 2 x 3 = 6

                                5! = 1 x 2 x 3 x 4 x 5 = 120

∴  n! = 1 x 2 x 3 x 4 x …… x n

        2! x

...read more.

Middle

    3

    4

    5

    6

  (n + 1)!

    2

    6

    24

   120

   720

   5040

    Sn

    1

    5

    23

   119

   719

   5039

Looking at the row of (n + 1)! and Sn, there is a constant difference of 1 between them. Therefore I added the row of Sn to the second table (in black).

When n = 1,                 (n + 1)! = 2        ;        Sn = 1

n = 2,                (n + 1)! = 6        ;        Sn = 5

.

        .

                         .

n = 6,                (n + 1)! = 5040;        Sn = 5039

        According to what I have discovered, Sn can be express mathematically in this way:

Sn = 1 x 1! + 2 x 2! + 3 x 3! + ...... + n x n! = (n + 1)! – 1

To prove that my expression is right, I am using mathematical induction to verify the given result:

Pn : 1 x 1! + 2 x 2! + 3 x 3! + ...... + n x n! = (n + 1)! – 1

Pk : 1 x 1! + 2 x 2! + 3 x 3! + ...... + k x k! = (k + 1)! – 1

If Pk+1 is true the result should be (k + 1 + 1)!– 1 = (k + 2)! – 1

Pk+1 : 1 x 1! + 2 x 2! + 3 x 3! + ...... + k x k! + (k + 1) x (k + 1)!

= (k + 1)! – 1 + (k + 1) x (k + 1)!

= (k + 1)! [(k + 1) + 1] – 1

        = (k + 1)! (k + 2) – 1

        = (k + 2)! – 1

∴ Pk+1 is true.

P1 :         LHS = 1 x 1! = 1

        RHS = (1 + 1)! – 1 = 1  

∴ P1 is true.

∴ Pn is true for all positive integers n.

Already I have derived the formula an = (n + 1)! – n! from the first table. But there is still another way to derive it just from the original formula:

                                an = n x n!

                                  = (n + 1 – 1) x n!

= (n + 1) x n! – n!

                                  = (n + 1)! – n!

Sn = a1 + a2 + a3 + a4 + a5 + …… + an

= (1 + 1)! – 1! + (2 + 1)! – 2! + (3 + 1)! – 3! + (4 + 1)! – 4! + (5 + 1)! – 5! + ......+ (n + 1)! – n!

=2! – 1! + 3! – 2! + 4! – 3! + 5! – 4! +6! – 5! + ...... + (n + 1)! – n!

...read more.

Conclusion

    1

    2

    3

    4

    5

    6

an : n x n!

    1

    4

    18

    96

   600

   4320

an+1 : (n + 1) x (n + 1)!

    4

    18

    96

   600

   4320

  35280

      (n + 1)!

    2

    6

    24

   120

   720

   5040

      (n + 2)!

    6

    24

   120

   720

   5040

  40320

       n!

    1

    2

    6

    24

   120

   720

cn : (n + 2)! – n!

    5

    22

   114

   696

   4920

  39600

T1 = c1 = 1

T2 = c1 + c2 = 5 + 22 = 27

T3 = c1 + c2 + c3 = 5 + 22 + 114= 141

T4 = c1 + c2 + c3 + c4 = 5 + 22 + 114 + 696 = 837

T5 = c1 + c2 + c3 + c4 + c5 = 5 + 22 + 114 + 696 + 4920 = 5757

T6 = c1 + c2 + c3 + c4 + c5 + c6 = 5 + 22 + 114 + 696 + 4920 + 39600 = 45357

       Tn

    5

    27

   141

   837

   5757

  45357

Looking at the column of (n + 1)!, (n + 2)! and Tn, when I add (n + 1)! to (n + 2)!, there is a constant difference of 3 between the sum and Tn. According to what I have found out, Tn can be express mathematically like this:

Tn = (1 + 2)! - 1! + (2 + 2)! – 2! + (3 + 2)! – 3! + (n + 2)! – n!

 = (n + 1)! + (n + 2)! – 3

Tn = c1 + c2 + c3 + c4 + c5 + ...... + cn-1 + cn

∵ cn = (n + 2)! – n!

Tn = (1 + 2)! - 1! + (2 + 2)! – 2! + (3 + 2)! – 3! + (4 + 2)! – 4! + (5 + 2)! – 5! + ...... + (n – 1 + 2)! – (n – 1)! + (n + 2)! – n!

  = 3! - 1! + 4! – 2! + 5! – 3! + 6! – 4! + 7! – 5! + ...... + (n + 1)! (n – 1)! + (n + 2)! – n!

At this step, I can see that many numbers cancel out each other except (-1!), (-2!), [(n + 1)!] and [(n + 2)!] as it goes on to the last term:

= (n + 1)! + (n + 2)! – 1! – 2!

∴ Tn = (1 + 2)! - 1! + (2 + 2)! – 2! + (3 + 2)! – 3! + (n + 2)! – n! = (n + 1)! + (n + 2)! – 3

In conclusion, throughout the investigation, I have used different methods to find out patterns of the sequences and successfully conjecture expressions for different sequences. Moreover, to prove the conjecture, I used not only by mathematical induciton, but also another method which I carried out for the last part. The 2 main conjectures I have made is:

Sn = (n + 1)! – 1

Tn = (n + 1)! + (n + 2)! – 3

And both of the expressions are true for all positive integers n.

...read more.

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