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Alone

01/01/2018 at 12:04
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4
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Let M=\(\dfrac{2.1+1}{\left(1^2+1\right)^2}+\dfrac{2.2+1}{\left(2^2+2\right)^2}+.......+\dfrac{2.2015+1}{\left(2015^2+2015\right)^2}\).Prove that:M<1




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    Lê Quốc Trần Anh Coordinator 02/01/2018 at 21:33

    M = \(1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{2015^2}-\dfrac{1}{2016^2}\)

    M = \(1-\dfrac{1}{2016^2}=1-\dfrac{1}{4064256}=\dfrac{4064255}{4064256}\)

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    FC Alan Walker 25/02/2018 at 14:34

    We have: \(\dfrac{2n+1}{\left(n^2+n\right)^2}=\dfrac{\left(n+1\right)^2-n^2}{n^2\left(n+1\right)^2}=\dfrac{1}{n^2}-\dfrac{1}{\left(n+1\right)^2}\)

    So: \(M=\dfrac{2.1+1}{\left(1^2+1\right)^2}+\dfrac{2.2+1}{\left(2^2+2\right)^2}+...+\dfrac{2.2015+1}{\left(2015^2+2015\right)^2}\)

                \(=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{2015^2}-\dfrac{1}{2016^2}\)

                \(=1-\dfrac{1}{2016^2}< 1\)

  • ...
    FA Liên Quân Garena 08/01/2018 at 21:57

    M = 1−122+122−132+...+120152−120162

    M = 1−120162=1−14064256=40642554064256

  • ...
    Lê Quốc Trần Anh Coordinator 02/01/2018 at 21:48
    So M < 1

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