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Math Help - Sequences, Series

  1. #1
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    Sequences, Series

    4. Prove that the terms of the sequence U_n = n^{2}-10n+27 are positive. For what value of n is U_n smallest?

    How do I do this?
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  2. #2
    Newbie gs.sh11's Avatar
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    Well first, n = 0 for U_n to be the lowest, because any negative numbers squared would turn out positive, 0 is smaller than any positive number, so to answer your question n = 0. Im not sure how to prove that they are positive, but I think it has something to do with squaring, anytime you square a negative it becomse positive, and -10 times any negative number will become positive.
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by Viral View Post
    4. Prove that the terms of the sequence U_n = n^{2}-10n+27 are positive. For what value of n is U_n smallest?

    How do I do this?
    First we need to complet the square

    n^2-10n+27=n^2-10n+25+2=(n-5)^2+2

    Since (n-5)^2 \ge 0 we know that

    (n-5)^2+2\ge 0+2=2>0 and since this is a parabola its mininum occurs at the vertex when n=5 you get 2.
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  4. #4
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    Can you expand on:

    <br /> <br />
(n-5)^2+2\ge 0+2=2>0<br />

    a bit please. I understood the completing the square bit fine.
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  5. #5
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by Viral View Post
    Can you expand on:

    <br /> <br />
(n-5)^2+2\ge 0+2=2>0<br />

    a bit please. I understood the completing the square bit fine.
    If you square any real number it is always non negative

    so (n-5)^2 is always bigger than or equal to zero.

    i.e the smallest it can ever be is 0 and that occurs when n=5

    so the smallest value your squence can take it 2.
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  6. #6
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    Thanks I got that now. How would I work out the first part of the question though? (the proof)
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