ប្រៀបធៀបចំនួនលោការីត


ប្រៀបធៀបចំនួនលោការីត

ប្រៀបធៀបចំនួនលោការីត

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One Response

  1. វិធីទីមួយ៖ (ប្រើលក្ខណៈគ្រឹះ) \forall n>1 យើងបាន
    + n+1>n\Leftrightarrow \dfrac{1}{n+1}<\dfrac{1}{n}\Leftrightarrow 1+\dfrac{1}{n+1}<1+\dfrac{1}{n}

    \Leftrightarrow \dfrac{n+2}{n+1}<\dfrac{n+1}{n}\Leftrightarrow log_n\Bigl(\dfrac{n+2}{1}\Bigr)n\Leftrightarrow log_{\dfrac{n+2}{n+1}}(n+1)1)

    \Leftrightarrow log_{n+1}\Bigl(\dfrac{n+2}{n+1}\Bigr)< log_n\Bigl(\dfrac{n+2}{n+1}\Bigr)\quad (2)

    តាម (1) & (2) \Rightarrow log_{n+1}\Bigl(\dfrac{n+2}{n+1}\Bigr)< log_n\Bigl(\dfrac{n+1}{n}\Bigr)

    \Rightarrow log_{n+1}(n+2)-log_{n+1}(n+1)< log_n(n+1)-log_nn

    \Rightarrow log_{n+1}(n+2)1;\ log_n(n+1)> log_{n+1}(n+2)

    ពីនេះ យក n=2011 យើងបានលំហាត់ត្រូវបានស្រាយបញ្ជាក់។

    វិធីទី២):

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