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# Investigating a sequence of numbers

Extracts from this document...

Introduction

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 is defined by

, , , …

From the pattern of different values of n in  above, I conclude that!

Part 2:

Let

If n=1

= were !

!

If n=2

were !, !

+ (2

If n=3

were !, !,!

+ (2

Part 3:

From Part 2, I know that:

!

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

Table 1.1:

 n 1 1 1 2 4 5 3 18 23 - - - - - - n ! ?

The same results of  from Table 1.

Middle

is true

1. If  is true, then

If k=k+1, then

Now,

=

=

=

=(k+2)!-1    (k+2)!

Thus is true whenever is true and  is true.

is true for all n

Part 5:

!, I use this formula to show that  is also true by simplifying  to equal !

! = (n+1)!-n!

= (n+1)n!-n!  (n+1)! = (n+1)n!

= n! (n+1-1)

= !

!

Now, I use  to device a direct proof for the expression of that I conjectured in Part 3.

!

From Part 3, I know that

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:

The conjecture of is proven

Part 6:

Conclusion

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

Graph 2.1

Part 9:

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

is:

+ (2+2)!-2! + … +

1. If n=1

LHS:

= (1+2)!-1!

= 5

RHS: (1+1)!(1+3)= 5

is true

1. If  is true, then

+ (2+2)!-2! + … +

If k=k+1, then

Now,

=

=

= (k+1)!(k+2)

= (k+2)!(k+4)

Thus is true whenever is true and  is true.

is true for all n

Conclusion:

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.

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

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