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Homework Help: In terms of n, 3, 7, 13, 27, 53, 107

  1. May 11, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the explicit formula c1=1, c2=2, cn+1 = (cn + cn-1)/2 ; n>2


    2. Relevant equations



    3. The attempt at a solution

    c1 = 1
    c2 = 2
    c3 = 3/2
    c4 = 7/4
    c5 = 13/8
    c6 = 27/16
    c7 = 53/32
    c8 = 107/64

    I know the bottom term is 2^n-2. I cannot find what the top is. If anyone sees it let me know thx.

    Also if you know general form of 1, 1, -1, -1, 1, 1, -1, -1 ....

    Thx a lot
     
  2. jcsd
  3. May 12, 2010 #2
    The numerator kind of looks like [tex]a_{n+2}=2a_{n}+a_{n+1}[/tex] to me, so it would be like this:
    c4 = (2*2) + 3 = 7
    c5 = (2*3) + 7 = 13
    c6 = (2*7) + 13 = 27
    etc, it only works when n > 2 which seems to fit into what you mentioned as well. Hope that helps a bit!
     
  4. May 12, 2010 #3

    lanedance

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    Homework Helper

    try multplying out explicitly for the first few terms - you should hopefully be able to pick up a pattern in terms of sums of powers of 2
    [tex]3 = 1+2[/tex]
    [tex]7 = 1 +2+2^2[/tex]
    and follow on from there...
     
  5. May 12, 2010 #4
    I was told you cannot use the terms before it in the equation, ie. cn-1*2 + (-1)^n
     
  6. May 12, 2010 #5

    lanedance

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    Homework Helper

    i'm not sure i know what you mean? try this though, write the c_n in term of a new variable a_n, for n>2
    [tex]c_1 = 1[\tex]
    [tex]c_2 = 2[\tex]
    [tex]c_3 = \frac{3}{2} = \frac{a_3}{2}[\tex]
    [tex]c_4 = \frac{7}{2^2} = \frac{a_4}{2^2}[\tex]

    so you want to find the a_n, with
    [tex]c_n = \frac{a_n}{2^{n-2}}[\tex]

    how about seeing if you can find a recursion relation for the a_n, using the original... will be similar to what refraction posted, but the form should help lead to a general solution.
     
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