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Math Help - induction - I got most of the proof, just cant finish

  1. #1
    Member pberardi's Avatar
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    induction - I got most of the proof, just cant finish

    Prove that for every natural number greater than or equal to 6, (n+1)^2 is less than or equal to 2^n

    Let S = {n in N s.t. n>=6, (n+1)^2 <= 2^n }

    Show 6 is in S 7^2 <= 2^7
    Assume n is in S
    Show n+1 is in S
    (n+2)^2 <= 2^n +1
    n^2 + 4n + 4 <= 2^n +1 HELP!
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  2. #2
    MHF Contributor
    Grandad's Avatar
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    Hello pberardi
    Quote Originally Posted by pberardi View Post
    Prove that for every natural number greater than or equal to 6, (n+1)^2 is less than or equal to 2^n

    Let S = {n in N s.t. n>=6, (n+1)^2 <= 2^n }

    Show 6 is in S 7^2 <= 2^7
    Assume n is in S
    Show n+1 is in S
    (n+2)^2 <= 2^n +1
    n^2 + 4n + 4 <= 2^n +1 HELP!
    n \in S \Rightarrow (n+1)^2 \le 2^n

    \Rightarrow n^2 + 2n+1 \le 2^n

    \Rightarrow n^2 + 4n+4 \le 2^n + 2n+3

    \Rightarrow (n+2)^2 \le 2^n + 2+2n+1

    \Rightarrow (n+2)^2 \le 2^n + n^2 + 2n + 1, since 2 < n^2 for n \ge 6

    \Rightarrow (n+2)^2 \le 2^n + (n+1)^2

    \Rightarrow (n+2)^2 \le 2^n +2^n, since (n+1)^2 \le 2^n

    \Rightarrow (n+2)^2 \le 2.2^n = 2^{n+1}

    Incidentally, to show 6 \in S, you should write 7^2 \le 2^6, not 2^7

    Grandad
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  3. #3
    Member pberardi's Avatar
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    Thanks for the clear proof and explanation sir.
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