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Series/Sequence Problem?

  1. Apr 14, 2008 #1
    [SOLVED] Series/Sequence Problem?

    I'm trying to figure out a formula for this sequence problem, rather than doing this over and over 1,000 times. Does anyone have a clue for the formula to find this? I've pasted the info and also the answer.

    Thanks for your time!
     

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  3. Apr 15, 2008 #2
    Write out the formula for a1, a2, and a3 without summing and find the pattern so that you can write a formula for an that does not refer to an-1. It should remind you of a simple type of sum you already know a shortcut for.
     
  4. Apr 15, 2008 #3

    HallsofIvy

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    {an} is defined, recursively, by a1= 4, [itex]a_{n+1}= a_n+ 4n[/itex]. Find a1000.

    The first thing I would do is start calculating a few values (hoping I won't have to go up to 1000!).

    a2= 4+ 4(1), a3[/sup]= 4+ 4(1)+ 4(2), a4[/sup]= 4+ 4(1)+ 4(2)+ 4(3).

    Hmmm, looks to me like an= 4(1+ 2+ 3+ ...(n-1)) so a1000= 4(1+ 2+ 3+ ... + 999). Can you find 1+ 2+ 3+ ...+ 999? It's an arithmetic sequence with common difference 1. Or you could use "Euler's method".
     
  5. Apr 15, 2008 #4

    tiny-tim

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    Another method: rewrite the equation as:

    an+1 - an = 4n;
    so an+2 - an+1 = 4(n+1).​

    Add them … what do you get … ? And then … ? :smile:
     
  6. Apr 15, 2008 #5
    Gentlemen, I'm totally stuck. I sat here for the last 1.5 hrs and tried to figure this one out. I think that I'm making it too difficult. I've looked at the sequence all the way up through a6 and I'm just not seeing the formula.

    I do know that to find the sum from 1 to 1000, would be (1000(1001))/2 or (n(n+1))/2. But I've tried finding the an like this also...dang I'm making this tuff aren't I?
     
    Last edited: Apr 15, 2008
  7. Apr 15, 2008 #6

    tiny-tim

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    Hi ae4jm! :smile:

    Humour me … what's an+2 - an? … what's an+3 - an? :smile:
     
  8. Apr 15, 2008 #7
    I got 12 and 24???
     
  9. Apr 15, 2008 #8

    tiny-tim

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    Hi ae4jm! :smile:

    Yes … that's a3 - a1 and a4 - a1.

    But you'll get more of an idea of what's going on if you answer the general question:

    what's an+2 - an? … what's an+3 - an? (in terms of n):smile:
     
  10. Apr 15, 2008 #9
    I believe that the first one is 6-4=2 and 7-4=3, right?
     
  11. Apr 15, 2008 #10
    does this look correct for
    Code (Text):
    [tex]a_{n}=4(\frac{n(n-1)}{2})+4[/tex]
     
  12. Apr 16, 2008 #11

    tiny-tim

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    :biggrin: Woohoo! :biggrin:

    And now, of course, simplify it to 2n(n - 1) + 4.

    Any questions? :smile:
     
  13. Apr 16, 2008 #12
    Thanks! That was a dandy, for me anyways! I'm sure this practice will help me recognize these sequences a little better in the future. Thanks for sticking in there with me.
     
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