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Combinatorics Question

  1. Oct 18, 2013 #1
    1) an = an-1 + 2n with a0 = 2
    The question is Use back-substitution to find a closed formula for the recurrence relations.
    I have an-4 + 2(n-3)+2(n-2)+2(n-1)+2n
    Then I have a1 + ... +2(n-5)+2(n-4)+2(n-3)+2(n-2)+2(n-1)+2n
    First of all, is my way right? If so, could you please help me find the recurrence relations?
    If no, could you tell I have done wrong and point me to the right direction?

    2) Find and solve a recurrence relation for pn the value of a stock market indicator that obeys the rule that the change this year from the previous year (that is, pn - pn-1) equals twice last year's
    change. Let p0 = 1; p1 = 4.
    I have a trouble to translate " the rule that the change this year from the previous year (that is, pn - pn-1) equals twice last year's" into recurrence relation. Can someone please help?
     
  2. jcsd
  3. Oct 19, 2013 #2

    LCKurtz

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    I presume you mean ##a_n = a_{n-1} +2n## with ##a_0=2##. I would write out the first few, taking care not to simplify, which might hide any patterns:

    ##a_1 = 2 + 2\cdot 1##
    ##a_2 = 2 + 2\cdot 1 + 2\cdot 2##
    ##a_3 = 2 + 2\cdot 1 + 2\cdot 2 + 2\cdot 3##

    Do you see any patterns or useful expressions you can use to get a nice closed formula for ##a_n##?
     
  4. Oct 19, 2013 #3
    The answer should be 2+n+n^2?

    Thanks
     
    Last edited: Oct 19, 2013
  5. Oct 19, 2013 #4
    Sounds to me like: [itex] (p_2-p_1)=2(p_1-p_0) [/itex] so, [itex] p_n-p_{n-1}=2(p_{n-1}-p_{n-2}) [/itex].
     
  6. Oct 19, 2013 #5

    LCKurtz

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    Can't you check it for yourself?
     
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