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Math algebra/induction

  1. Sep 30, 2012 #1
    1. The problem statement, all variables and given/known data


    The induction question is. for all natural n, n4 <= 4n + 17


    Base case: 0 Works, since 0 < 1 + 17 then,
    I assume that for all n in natural, n4 <= 4n + 17 holds.


    Now I believe I need to show that, 4(n4) <= 4(4n + 17)
    that is, 4n+1 + 17 >= (n+1)4
    To do so, I prove, 4n4 >= (n+1)4,
    which proves that 4n+1 + 17 >= (n+1)4


    How would I prove.. 4n4 >= (n+1)4 = n4 + 4n3 + 6n2 + 4n + 1

    This step is in the middle of my induction proof and it is neccesary part of my induction step.

    How would I go about doing this?
    Some easier version similar to this deals with power of 2 or n, which seems rather simple. but, this one I am having hard time.
    Help is much appreciated.

    2. Relevant equations



    3. The attempt at a solution

    I tried starting from n4 = n4 and start adding things to both sides but the onlything I can add to left is n4 so I am not entirely sure how to go about doing this type of math. please some tricks and help is appreciated.
     
    Last edited: Sep 30, 2012
  2. jcsd
  3. Sep 30, 2012 #2

    SammyS

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    Please state the whole problem as it was given to you.

    You're more likely to get help that way.

    Also, at least sketch out your solution up to the point you're having trouble. It's hard for us to guess what you're trying to accomplish otherwise.
     
  4. Sep 30, 2012 #3
    Agreed, for example this expression certainly isn't true for n=1, so for the induction to take place, where are you starting from?
     
  5. Sep 30, 2012 #4

    I added the full question with my understandings.
     
  6. Sep 30, 2012 #5
    sorry I fixed the problem. it's >= instead of >
    thanks,
     
  7. Sep 30, 2012 #6
    But you've changed the whole question since you last asked it. It is now a mess. Unless you yet again change it.

    And n4+1 is not the same as 4n4 which is how you have it written here, before you change it again!
     
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