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Prove 2^n > 1/2(n+1)² is true help!

  1. Oct 19, 2006 #1
    i have understood step one (the easiest, hehehe)
    If n=4, then LHS (left hand side) = 16
    RHS (right hand side) = 12.5
    16>12.5
    Hence P4 is true....
    but then when it comes to step 2, im confused, im used to proving sequences by mathematical induction, and, here, there doesnt seem to be a sequence in the first place.....!
     
  2. jcsd
  3. Oct 19, 2006 #2

    radou

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    [tex]2^n > \frac{1}{2}(n+1)^2[/tex]

    For n = 1, you have 2 > 2. :smile:
     
  4. Oct 19, 2006 #3
    Sorry, n>3
     
  5. Oct 19, 2006 #4

    radou

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    Ok, not sure about this one, but, as you already showed, it is true for n = 4. Further on, we assume it is true for some k:
    [tex]2^k > \frac{1}{2}(k+1)^2[/tex] ...(1)

    We have to proove that it is true for k + 1:
    [tex]2^{k+1} > \frac{1}{2}(k+2)^2 = \frac{k^2}{2}+ 2k + 2[/tex]

    So, we can multiply both sides of (1) with 2 to obtain:
    [tex]2^k \cdot 2 > \frac{1}{2}(k+1)^2\cdot 2 \Rightarrow[/tex]
    [tex]2^{k+1} > (k+1)^2 = k^2+2k+2[/tex], which implies
    [tex]2^{k+1}>\frac{k^2}{2}+ 2k + 2[/tex].
     
  6. Oct 19, 2006 #5
    hmm, why are you multiplying both sides by two?
     
  7. Oct 19, 2006 #6

    radou

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    To obtain [tex]2^{k+1}[/tex] on the left side. Anyway, as I said, I'm not a hundred percent sure about this one, so maybe better wait for someone else to comment. :biggrin:
     
  8. Oct 19, 2006 #7
    hmm...if you add 2 on one side, you have to add it on the other....all right ill wait, although i dont think anyone will comment, ive been posting mathematical induction questions, and it's quite rare when people do actually answer....
     
  9. Oct 19, 2006 #8

    radou

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    What do you mean by 'add 2 on one side'? Each side was multiplied with 2.
     
  10. Oct 19, 2006 #9
    sorry i meant multiply!
    when we let k=k+1, isnt that only applicable for a sequence, here we're dealing with something completely different, there's no sequence....we just have to prove it...im really confused!!
     
    Last edited: Oct 19, 2006
  11. Oct 19, 2006 #10

    radou

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    Well, you shouldn't think of any sequences. It's the principle of mathematical induction. We assumed that the statement was true for some natural number k, and we prooved that it is true for k+1. Therefore, we conclude that the statement is true for every natural number n > 3.
     
  12. Oct 19, 2006 #11
    ok i see what you are coming at, but i dont know whether the multiplication part is correct, although at the end Pn happens to be true (if we are to use your method: 2^k+1>1/2([k+1]+1)). Usually, in mathematical induction, you add 2^k+1 to the RHS because you are proving that the sum is applicable for the k+1th term....multiplying two to either sides does not seem logical? (that's only my point of view).
     
  13. Oct 19, 2006 #12

    radou

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    There is no 'logic' in that sense here. You're talking about forms of induction proofs which apply for sequences, sums, etc. Maybe you need more examples - for instance, look at example 3 in this link: http://library.thinkquest.org/10030/11matind.htm.
     
  14. Oct 19, 2006 #13
    yeah that's helpful although they go into too much detail! your's makes sense, thanks radou! take care.
     
  15. Oct 27, 2006 #14
    how does 2^{k+1} > (k+1)^2 = k^2+2k+2 imply this 2^{k+1}>\frac{k^2}{2}+ 2k + 2 ? I think you skipped a step?
     
  16. Oct 27, 2006 #15

    radou

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    I didn't. [tex]k^2+2k+2 > \frac{k^2}{2}+2k+2[/tex]. If you have a relation b1 > b2, then the relation a > b1 implies a > b2, too.
     
  17. Oct 30, 2006 #16
    all right, sorry for the late reply. once again, thank you radou!
     
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