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Math Help - Induction proof...I'm stuck!

  1. #1
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    Exclamation Induction proof...I'm stuck!

    Hopefully someone can help me with this..I'm nearly done.

    I'm in the induction step of the proof & here's what I have:

    "So suppose that P(n) holds for n an integer and n>=5, we have that
    2^n > n^2.
    We wish to show that P(n+1) holds, that is that
    2^(n+1) > (n+1)^2. This can be rewritten as
    2(2^n) > n^2 +2n +1.
    So, by our induction hypothesis and our Lemma (stated & proved earlier before this proof) we have
    2(2^n) > 2^n > n^2 > 2n+1."

    ....This is where I'm stuck...is there a property that allows me to then just state "Thus 2(2^n) > n^2 +2n +1"?
    Thanx for any feedback you provide!
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  2. #2
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    2^{n+1}=2^n+2^n>n^2+2n+1\rightarrow 2^n>2n+1 using the hypothesis. This should be proved for n\ge 5: Suppose that 2^n>2n+1; we show that 2^{n+1}>2(n+1)+1. This is equvivalent to: 2^n+2^n>2n+3\rightarrow 2^n>2 using the hypothesis. This is true even for n\ge 1. QED
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  3. #3
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    Are you using the fact that 2^(n+1) = 2^n + 2^n ? I started to use this in my proof then realized that although this is true for a base 2 it is not true for a base > 2 (i.e. 3^(n+1) does not = 3^n + 3^n...
    3^(n+1)=3(3^n).
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  4. #4
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    You know (used it) that 2^{n+1}=2\cdot 2^n. If you imagine that 2^n is an apple then 2\cdot 2^n is two apples so it apple+apple. Hope you understand it.
    And yes of course 3^n+3^n=2\cdot 3^n.
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  5. #5
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    Quote Originally Posted by james_bond View Post
    You know (used it) that 2^{n+1}=2\cdot 2^n. If you imagine that 2^n is an apple then 2\cdot 2^n is two apples so it apple+apple. Hope you understand it.
    And yes of course 3^n+3^n=2\cdot 3^n.
    Thanks James Bond!
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