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Questions about using iteration to express a recurisve sequence

  1. Nov 24, 2006 #1
    Hello everyone, i'm having some issues on a few problems and i'm not sure if i did the first one i'm going to post right, any help or clarifcation would be great.

    The directions are: In eavch of 3-15 a sequence is defined recursively. Use iteration to guess an explicit formula for the sequence.

    and here is my work:

    For the last one, I see that the 2 is increasing at the same rate as the 3^k, but i have no idea about the pattern, i'm not sure what technique i'm suppose to use, i know they said "guess" a formula, but what will help me make an educated guess? Of course there are some easy ones like n! that i can recongize but these arn't so easy for me to see whats going on.

    I can see i can factor out a 2 and a 3^2 from p_4 but i don't see how that is going to help me much, getting:
    [tex]2^4 + 3^4 + 2*3^2(1 + 3)[/tex]

  2. jcsd
  3. Nov 25, 2006 #2
    You made a bit of an arithmetic mistake: that should be [tex]2^2 3^2[/tex], not [tex]2^1 3^2[/tex]. If you continue on I think you'll notice that [tex]p_n = 2^n + 3^n + \left( 2^{n-2} 3^2 + 2^{n-3} 3^3 +...+ 2^1 3^{n-1} \right)[/tex]. That stuff in parentheses is a geometric series.
  4. Nov 26, 2006 #3
    Thanks durt!
    I'm alittle confused on how you got some of the terms of n
    [tex]p_n = 2^n + 3^n + \left( 2^{n-2} 3^2 + 2^{n-3} 3^3 +...+ 2^1 3^{n-1} \right)[/tex]

    But the parts of it I don't understand is how do you know after looking at a few terms, like a_1 to a_4 how you figure out which exponents go to a certian power, such as [tex]2^{n-2} 3^2 + 2^{n-3} 3^3 +...+ 2^1 3^{n-1}[/tex] ?

    I see that [tex]2^n + 3^n[/tex] would be just n because they are increasing at the same time, but from there whats the process to break that down into the other powers of n?


    Also I could replace that geometric series with the following sum right?
    First term is [tex]2^{n-2} 3^2[/tex]
    ratio is: 3/2 I believe because 2 is decreasing and 3 is increasing as you progress
    term after the last term is: [tex]3/2* 2^1 3^{n-1}[/tex] = [tex]3^{n-2}[/tex]

    so now applying a formula:
    [mythical next term - first term]/(ratio-1)
    = ( [tex]3^{n-2}[/tex] - [tex]2^{n-2} 3^2[/tex] )/(3/2 -1)
    Last edited: Nov 26, 2006
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