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Find the general formula

  1. Apr 22, 2006 #1
    Hi I have many problems trying to find the general formula and the demonstration by induction.
    Let
    [tex] A_1 = 3[/tex]
    [tex] A_2 = 7[/tex]
    [tex] A_3 = 13[/tex]
    A_n+3 = 3A_n+2 - 3A_n+1 + A_n
    I could only find this.
    A_n+1 = A_n + 2n
     
    Last edited: Apr 22, 2006
  2. jcsd
  3. Apr 22, 2006 #2

    Hurkyl

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    What sort of class are you in?


    How did you arrive at this? It can't be right, though: it doesn't work for n=1 or n=2.

    Oh, I bet you meant [itex]A_n = A_{n-1} + 2n[/itex], or [itex]A_{n+1} = A_n + 2(n+1)[/itex]. (Incidentally, you could try proving this formula by induction, to make sure you're on the right track)


    One trick that's often useful is to not do arithmetic. If [itex]A_n = A_{n-1} + 2n[/itex], then write [itex]A_2 = 3 + 2\cdot2[/itex] instead of [itex]A_2 = 7[/itex].
     
  4. Apr 22, 2006 #3
    sorry im in my way of learning Latex.
    I mean:
    [itex]A_n = A_{n-1} + 2n[/itex]
    And the formula given in the problem is:
    [itex]A_{n+3} = 3A_{n+2} - 3A_{n+1} + A_n[/itex]
    With the three A1 A2 A3 defined.
    Ive been trying but I couldnt find the general formula.
     
    Last edited: Apr 22, 2006
  5. Apr 22, 2006 #4

    VietDao29

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    So far so good.
    Now, you can follow Hurkyl's suggestion:
    A1 = 3
    A2 = A1 + 2 . 2 = 3 + 2 . 2
    A3 = A2 + 2 . 3 = 3 + 2 . 2 + 2 . 3
    A4 = 3 + 2 . 2 + 2 . 3 + 2 . 4
    A5 = 3 + 2 . 2 + 2 . 3 + 2 . 4 + 2 . 5
    A6 = 3 + 2 . 2 + 2 . 3 + 2 . 4 + 2 . 5 + 2 . 6
    ...
    An = 3 + 2 . 2 + 2 . 3 + 2 . 4 + 2 . 5 + 2 . 6 + 2 . 7 + ... + 2 . n
    Now, do you see anything that can be factored out?
    Can you go from here? :)
     
  6. Apr 22, 2006 #5
    For
    [itex] A_n = 3 + \sum\limits_{i = 2}^n 2i [/itex]
    then
    [itex] A_1 = 3 [/itex]
    Is this correct?
     
    Last edited: Apr 23, 2006
  7. Apr 22, 2006 #6

    Hurkyl

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    If you can prove it by induction, it's correct. :smile:
     
  8. Apr 23, 2006 #7
    Induction:
    1)
    [itex] A_1 = 3 + \sum\limits_{i = 2}^1 2i = 3 [/itex]
    (Is this OK?)
    2) I had this result :[itex]A_{n+1} = A_n + 2(n+1)[/itex]

    With [itex] A_n = 3 + \sum\limits_{i = 2}^n 2i [/itex] =>
    [itex]A_{n+1} = 3 + \sum\limits_{i = 2}^{n+1} 2i[/itex] (the upper index of the summatory in this case is (n+1))
    => [itex] A_{n+1} = 3 + \sum\limits_{i = 2}^n 2i + 2(n+1)[/itex]
    => [itex] A_{n+1} = A_n + 2(n+1) [/itex]

    Is this correct?
     
    Last edited: Apr 23, 2006
  9. Apr 23, 2006 #8

    Hurkyl

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    Well, you've done it backwards!

    [tex] A_{n+1} = A_n + 2(n+1) [/tex]

    is the statement you know to be true, whereas

    [tex]A_{n+1} = 3 + \sum\limits_{i = 2}^{n+1} 2i[/tex]

    is the statement you were trying to prove.

    (given the assumption that [itex] A_n = 3 + \sum_{i = 2}^n 2i [/itex])



    (Actually, I'm assuming you've already given a proof of [itex] A_{n+1} = A_n + 2(n+1) [/itex] is correct given the original recurrence relation. If you have not yet done so, then you should work with the original recurrence)
     
  10. Apr 23, 2006 #9
    Yes is true.I still have to prove first [tex] A_{n+1} = A_n + 2(n+1) [/tex]
     
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