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Progression proof

  1. Dec 24, 2012 #1
    1. The problem statement, all variables and given/known data
    we have u(1)=1
    u(n+1)=[itex]\frac{1}{16}[/itex](1+4u(n)+[itex]\sqrt{1+24u(n)}[/itex])

    v(n)^2=1+24u(n)

    Prove that [itex]\foralln[/itex][itex]\inN*[/itex]: v(n+1)-3=[itex]\frac{1}{2}[/itex](v(n)-3)
    2. Relevant equations



    3. The attempt at a solution
    by working on v(n+1)-3 and replacing it by [itex]\sqrt{1+24u(n+1)}[/itex] didn't get me anywhere just a bunch of calc
    also i made 6u(n+1)=[itex]\frac{1}{16}[/itex](v(n)+[itex]\frac{1}{2}[/itex])^2+[itex]\frac{19}{64}[/itex] but it didn't work for me , also just to mention the progression is not arithmetic or anything it's just (u(n)):n[itex]\geq1[/itex]
    PS: the "^2" means squared
     
    Last edited: Dec 24, 2012
  2. jcsd
  3. Dec 24, 2012 #2

    haruspex

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    Have you tried induction? Btw, there's a parenthesis missing in un+1=(1/16)(1+4un+√(1+24un). Should it be un+1=(1/16)(1+4un)+√(1+24un)?
     
  4. Dec 24, 2012 #3
    sorry forgot the parenthesis , using induction will make things worse ethe problem is the square root it's causing the problem here..
     
  5. Dec 24, 2012 #4

    haruspex

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    Induction doesn't look too bad. Don't get tangled up with square roots - work with the squared-up form. Assume true up to
    vn-3=(vn-1-3)/2
    i.e.
    vn=(vn-1+3)/2
    Now work with vn+12 and see if you can show it equal to ((vn+3)/2)2. (Work from both ends.)
     
  6. Dec 25, 2012 #5
    Thanks solved it.
     
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