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Math Help - PRoof by induction: Inequality

  1. #1
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    Cool PRoof by induction: Inequality

    I need to prove that for n>= 7, 3n^2>17n+10

    I know the first steps proving for n=7, then assuming for k, but once i need to prove for (k+1) i am stumped.
    Thank you for your time and help!

    Rob
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  2. #2
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    Re: PRoof by induction: Inequality

    Base case: For n = 7, you have
    3 n2 = 3 * 49 = 147
    17n + 10 = 17 * 7 + 10 = 129
    So, 3n2>17n+10

    Induction: Assume the inequality holds for n = k > 7. That is, assume that 3k2>17k+10. We need to prove that the inequality holds for n = k +1. That is, we need to prove that 3(k+1)2>17(k+1)+10.
    But 3(k+1)2 = 3k2 + 6k + 3. We know from our assumption 3k2>17k+10. So, we get 3(k+1)2 = 3k2 + 6k + 3 > 17k + 10 + 6k + 3 > 17k + 17 + 10 = 17(k+1) + 10. (Note: 6k+3 > 17 when k > 7.)
    Thanks from rchrist3
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  3. #3
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    Re: PRoof by induction: Inequality

    thank you for your help! now i worked it all out and it went like this:
    17k+10+6k+3>17k+17+10
    : 23K+13>17k+27
    : 23k>17k+14
    :6k>14
    : k> 2.3

    i feel like i did something wrong... or this is right because if it works for k>2.3 we know it is true bc out k is >= 7?
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  4. #4
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    Re: PRoof by induction: Inequality

    Your reasoning is correct.
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