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Mathematical Induction

  1. Jan 7, 2007 #1
    Here's my problem: 1 * 2 + 2 * 3 + 3 * 4 + . . . + n( n+ 1) = n(n + 1)(n + 2) / 3


    3. The attempt at a solution I know that there is a Basis step and Induction step. For the Basis step I have the following but don't know if I'm on the right track: Basis Step 1 * 2 = 1 * 2 * 3 / 3 , which is true

    I get lost at the Inductive step.

    Any help would be appreciated
     
    Last edited: Jan 7, 2007
  2. jcsd
  3. Jan 7, 2007 #2
    Welcome to PF,Bucs.

    Which part of the induction step do you have a problem with?
     
  4. Jan 7, 2007 #3

    cristo

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    Your basis step is right. Where, specifically, do you have problems?

    edit: neutrino beat me to it!
     
  5. Jan 7, 2007 #4
    Where I get stuck is what I should be inputing into the statement to verify it.

    1*2*3 + n(n+1) = n(n + 1)(n + 2) / 3 +

    I'm stuck on how to work the rest of the problem out.
     
  6. Jan 7, 2007 #5

    You have to now prove that the above statement holds for n = k+1, assuming that is true for some positive k.

    1.2+2.3+...k(k+1) + (k+1)((k+1)+1)

    Now, you have to show it takes the form n(n + 1)(n + 2) / 3, for n = k+1.
     
  7. Jan 7, 2007 #6
    I'm not sure that I'm following you - So it would look like this

    n(n + 1)(n + 2) / 3 + k(k+1)

    n(n + 1)(n + 2) / 3 + k(k+1)/3 / 3
     
  8. Jan 7, 2007 #7
    1.2+2.3+...k(k+1) + (k+1)((k+1)+1) ---> This is for k+1

    Since we've assumed that it holds true for some arbitrary k greater than 1, we can write

    1.2+2.3+...k(k+1) = k(k + 1)(k + 2) / 3 (1)

    Now,
    1.2+2.3+...k(k+1) + (k+1)((k+1)+1) = k(k + 1)(k + 2) / 3 + (k+1)((k+1)+1) (refer(1))

    Simpligy the right-hand side to (k+1)(k+2)(k+3)/3, which is what you would get if you put n = K+1 in n(n + 1)(n + 2) / 3
     
  9. Jan 7, 2007 #8
    Thanks for all of your help neutrino - I think the whole Inductive step is what is really fowling me up - my algebra is poor at best.
     
  10. Jan 7, 2007 #9
    Can I ask you 1 more question though?

    2n + 1 < 2n, n = 3, 4 …..
    Basis Step (n = 3) - 2(3) = 6 + 1 = 7 < 23 = 8 so we Assume true for n

    Am I on the right track?
     
  11. Jan 7, 2007 #10
    :confused:

    Are you sure that's the question?

    and this whole part...

    2(3) = 6 + 1 = 7 < 23 = 8

    doesn't make sense.
     
  12. Jan 7, 2007 #11
    I thought by plugging in either 3 or 4 as n that it would be true. I have to verify the inequality
     
  13. Jan 7, 2007 #12

    cristo

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    This cannot be the correct question, since 2n+1>2n for all n!!
     
  14. Jan 7, 2007 #13
    14. 2n + 1 < 2n, n = 3, 4 …..

    It should be less than or equal to - it just didn't come through when I pasted it.
     
  15. Jan 7, 2007 #14

    cristo

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    Should it not read something like [itex]2n+1\leq 2^n[/itex] for n=3,4,...?
     
  16. Jan 7, 2007 #15
    See - I really am bad at this stuff - the way you have written it is correct.

    From my above post I guess my Basis Step is wrong?
     
    Last edited: Jan 7, 2007
  17. Jan 7, 2007 #16

    cristo

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    Well, your basis step: It is true for n=3 so assume true for n=k by the inductive hypothesis. So, we know that [itex]2n+1\leq 2^n[/itex], for n=3,...,k, and we want to show true for n=k+1. So, we have [itex]2(k+1)+1=2k+2+1[/itex]. Can you invoke the fact that we know is true (i.e. it holds for n=k) into this equation?
     
  18. Jan 7, 2007 #17

    HallsofIvy

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    If you think that 2n, when n= 3, equals 23, this may be hopeless. Do you know what "2n" means?
     
  19. Jan 7, 2007 #18
    I'm not sure what you mean by "it holds for n=k"
     
  20. Jan 7, 2007 #19

    cristo

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    You have proven by your inductive hypthesis that [itex]2k+1\leq 2^k[/itex]
     
  21. Jan 7, 2007 #20
    That's all that's to that problem? I thought it would be much more difficult than that.
     
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