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Math Help - Induction

  1. #1
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    Induction

    See reply below, thanks!
    Last edited by ff4930; October 5th 2007 at 02:29 PM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ff4930 View Post
    Im confused at this problem, any help would be appreciated.

    Let
    <br />
a_{1} = a<br />

    <br />
a_{n} = 2a_{n-1} + 1<br />

    Prove by induction that
    <br />
a_{n} = 2^(n) - 1<br />

    Can a be any integer?
    if I put a_{1} = 5
    then a_{1} = 5 = 5
    if I let n = 2
    then
      a_{2} = 2(5)+1 = 11 != 2^(n) - 1 = 2^2-1 = 3


    Thats the confusion, please help.
    that's true, a_1 = 5 does not work for this. are you sure you copied the question correctly? have you left any information or conditions out? i think it works for a_1 = 1 though, then the proof would be easy, in fact, i think we can show that a_1 = 1 is the only plausible first term
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  3. #3
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    This is an print out review exam problem, I just checked online and she has took this problem out. I had the older version. I guess my prof. saw the problem, sorry about that.

    If you guys can help me on my second most frustrated Induction, it be great.

    Use induction to prove that n! < n^n whenever n is a positive integer greater than 1.

    Base case: n = 2
    2! = 2 < 2^2 = 4 - checks!

    Assume n! < n^n, n = k.

    Prove that n! < n^n when n = k+1

    (k+1)! < (k+1)^(k+1)..thats how far I got, please help.
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  4. #4
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    Say that k > 1\quad \& \quad k! < k^k .
    Look at this.
    \left( {k + 1} \right)! = \left( {k + 1} \right)\left( {k!} \right) < \left( {k + 1} \right)\left( {k^k } \right) < \left( {k + 1} \right)\left( {\left[ {k + 1} \right]^k } \right).

    Can you finish?
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  5. #5
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    Quote Originally Posted by Plato View Post
    Say that k > 1\quad \& \quad k! < k^k .
    Look at this.
    \left( {k + 1} \right)! = \left( {k + 1} \right)\left( {k!} \right) < \left( {k + 1} \right)\left( {k^k } \right) < \left( {k + 1} \right)\left( {\left[ {k + 1} \right]^k } \right).

    Can you finish?
    I'm sorry I'm just so bad in Induction. I can keep staring and I still wont see it.

    Did you add (k+1) to both sides?
    I lost you when you got up to \left( {k + 1} \right)\left( {\left[ {k + 1} \right]^k } \right)
    Could you explain on how you got that please.
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  6. #6
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    It is really so simple!
    k + 1 > k\quad  \Rightarrow \quad k^k  < \left( {k + 1} \right)^k
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  7. #7
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    wrong post sorry
    Last edited by darken4life; October 6th 2007 at 07:18 AM.
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