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Finding general term of a sequence

  1. Sep 13, 2014 #1
    1. The problem statement, all variables and given/known data

    Find the general term of

    0, 4, 22, 118, 718, 5038, 40318

    if a1 = 0, a2 = 4 and so on.


    3. The attempt at a solution

    I have tried getting the differences between them, even going 4 levels deep, but that isn't working obviously, I tried seeing if the ratios are the same, but they aren't. I really don't know what to do, I just need a hint to get this problem going. Thanks for your time!
     
  2. jcsd
  3. Sep 14, 2014 #2

    phinds

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    The next term is 362878. Doesn't seem very hard, I guess you just have to see the trick.
     
  4. Sep 14, 2014 #3

    HallsofIvy

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    There are, of course, an infinite number of such sequences. There is, for example a unique polynomial, of degree 6 or less, passing through those 7 points. To get that you would have to take differences "6 levels deep".
     
  5. Sep 14, 2014 #4

    nmr

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    I'm not the original poster but still interested in solving this problem. Following your hint, the following is the expression I ended up with.

    A_n = A_(n-1) * (n+1) + 2(n+1)

    How do I go about resolving A_(n-1) to a function/expression in terms of n?
     
  6. Sep 14, 2014 #5

    phinds

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    You have to take an assumed starting point, I think. I certainly did, but maybe someone with more math experience can give you an answer.

    By the way, I obviously could have just posted the answer myself, but I thought I'd give the OP another shot at it, knowing that it isn't hard. I think you showed poor form in not doing the same.

    The point of this forum is to help others figure things out, NOT to spoon feed them answers.
     
  7. Sep 14, 2014 #6
    Going six levels deep would only give me one number left, I am thinking you would need at least 3 to see if the differences are the same. By the way, for the last value I got 18806. I am kind of stumped again. Yes I typed the sequence into wolfram alpha and got answers with factorials everywhere, I know I wouldn't have got that on my own.
     
    Last edited: Sep 14, 2014
  8. Sep 14, 2014 #7

    Ray Vickson

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    It is even poorer form to post an incorrect solution, as the previous OP has done.
     
  9. Sep 14, 2014 #8
    I figured it out, (N+1)! - 2, thank you all!
     
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