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SAT/ GCSE-Level Recurrence Relation Problem

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  1. Jul 16, 2011 #1
    1. The problem statement, all variables and given/known data
    Hi! This is my first time on the site. I look forward to working with everyone…but hopefully not too much, assuming I‘m learning things correctly. :P

    My question pertains to Recurrence Relations, so here it goes…

    Foreword: The text book I’m using actually supplies the answer to the question, so I already have a point of reference, but my attempt does not match up with the answers. I believe my approach is essentially correct, as it has yielded the correct answers for a similar question beforehand. Answer is: 1, 3, 7, 17, 41

    Please note that I am beginning the question from u3, as we already have the values for u1 and u2.


    2. Relevant equations

    Q. Find the first five terms of the sequence:

    u1 = 1, u2 = 3 and un = 2un-1 + un-2


    3. The attempt at a solution

    Attempt:

    Solve un+1 where un = 3un-1 - un-2
    => 3u(n+1)-1 - u(n+1)-2

    Begin by substituting 3 (i.e. u2) for un:
    If n = 1 then u3 = 2((3+1) - 1) + ((3+1) -2) => 2(4-1) + (4-2) => 6 + 2
    Ans.: u3 = 8... but should be 7!!!

    Proceeding with u3 as 7, not 8...

    If n = 2 then u4 = 2((7+1) -1) + ((7+1) -2) => 2(8-1) + (8-2) => 14 + 6
    Ans.: u4 = 20... But should be 17!!!

    Note, I am omitting solution of u5 for brevity’s sake.

    I‘m sure the answer is staring me in the face, but I just can’t seem to figure it out!
    Can anyone help?

    Thanks.
     
    Last edited: Jul 16, 2011
  2. jcsd
  3. Jul 16, 2011 #2

    hunt_mat

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    Not too sure what you're going here but let's calculate [itex]u_{3}[/itex] from the recurrence relation.
    [tex]
    u_{3}=2u_{2}+u_{1}=2\times 3+1=7
    [/tex]
    Working for [itex]u_{4}[/itex]
    [tex]
    u_{4}=2u_{3}+u_{2}=2\times 7+3=17
    [/tex]
     
  4. Jul 16, 2011 #3
    Woah, that was easier than I was making it! Thank you.

    One final question though, why is the value of u1 subbed into un-2 and u2 into un-1?
     
  5. Jul 16, 2011 #4

    hunt_mat

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    you're finding n=3, so n-1=2 and n-2=1.
     
  6. Jul 16, 2011 #5
    Thank you very much. You've really helped me out!
     
  7. Jul 16, 2011 #6

    hunt_mat

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    it's why I help here.
     
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