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Math Help - Need help with the algebra in an induction proof

  1. #1
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    Need help with the algebra in an induction proof

    Hi, I've been working on this problem for quite a while now and I just can't seem to get the algebra working for it. Any help is appreciated.

    Determine the exact set of natural numbers n for which the inequality 2^n>=(n+1)^2 holds (1).

    I have already found that this is true for n>=6 from intuition and plugging stuff in, but I need to prove this by induction.

    Base case: n=6 holds for both sides

    Induction step: I need to prove that (1) holds for n=k+1.
    I tried plugging k+1 in for the left side, yielding 2^(k+1) = 2^k*2^1>=2*(k+1)^2 (from (1)) But this gets me nowhere as my ultimate goal is to arrive at 2^(k+1)>=((k+1)+1)^2
    Any ideas?
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  2. #2
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    Hello morbius27
    Quote Originally Posted by morbius27 View Post
    Hi, I've been working on this problem for quite a while now and I just can't seem to get the algebra working for it. Any help is appreciated.

    Determine the exact set of natural numbers n for which the inequality 2^n>=(n+1)^2 holds (1).

    I have already found that this is true for n>=6 from intuition and plugging stuff in, but I need to prove this by induction.

    Base case: n=6 holds for both sides

    Induction step: I need to prove that (1) holds for n=k+1.
    I tried plugging k+1 in for the left side, yielding 2^(k+1) = 2^k*2^1>=2*(k+1)^2 (from (1)) But this gets me nowhere as my ultimate goal is to arrive at 2^(k+1)>=((k+1)+1)^2
    Any ideas?
    Suppose that P(n) is the propositional function: 2^n >(n+1)^2

    Then P(n)
    \Rightarrow 2^n\times 2 > 2(n+1)^2
    \Rightarrow 2^{n+1} > 2n^2+4n + 2
    =n^2 + 4n + 4 +(n^2-2)

    >(n+2)^2, since n^2-2 >0 when n\ge 6
    \Rightarrow 2^{n+1} > ([n+1]+1)^2
    So P(n) \Rightarrow P(n+1)

    Grandad
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