Bài tập nâng cao · Bài 7
Bài tập nâng cao Lôgarit · Bài 7
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$\log_{2018}2019+2^{2}\log_{\sqrt{2018}}2019+3^{2}\log_{\sqrt[3]{2018}}2019+...+n^{2}\log_{\sqrt[n]{2018}}2019=1010^{2}.2021^{2}\log_{2018}2019$
$\log_{2018}2019+2^{2}\log_{\sqrt{2018}}2019+3^{2}\log_{\sqrt[3]{2018}}2019+...+n^{2}\log_{\sqrt[n]{2018}}2019=1010^{2}.2021^{2}\log_{2018}2019$
Xem lời giải
Lời giải
$\log_{2018}2019+2^{2}\log_{\sqrt{2018}}2019+3^{2}\log_{\sqrt[3]{2018}}2019+...+n^{2}\log_{\sqrt[n]{2018}}2019=1010^{2}.2021^{2}\log_{2018}2019$
$\Leftrightarrow \log_{2018}2019+2^{3}\log_{2018}2019+3^{3}\log_{2018}2019+...+n^{3}\log_{2018}2019=1010^{2}.2021^{2}\log_{2018}2019\Leftrightarrow \left( 1+2^{3}+3^{3}+...+n^{3} \right)\log_{2018}2019=1010^{2}.2021^{2}\log_{2018}2019$
$\Leftrightarrow 1+2^{3}+3^{3}+...+n^{3}=1010^{2}.2021^{2}$
$\Leftrightarrow \left( 1+2+...+n \right)^{2}=1010^{2}.2021^{2}$
$\Leftrightarrow \left[ \frac{n\left( n+1 \right)}{2} \right]^{2}=1010^{2}.2021^{2}$
$\Leftrightarrow \frac{n\left( n+1 \right)}{2}=1010.2021$
$\Leftrightarrow n^{2}+n-2020.2021=0$
$\Leftrightarrow \left[ \begin{array}{l} n=2020 \\ n=-2021\,\,\left( \ell \right) \end{array} \right.$
$\Leftrightarrow \log_{2018}2019+2^{3}\log_{2018}2019+3^{3}\log_{2018}2019+...+n^{3}\log_{2018}2019=1010^{2}.2021^{2}\log_{2018}2019\Leftrightarrow \left( 1+2^{3}+3^{3}+...+n^{3} \right)\log_{2018}2019=1010^{2}.2021^{2}\log_{2018}2019$
$\Leftrightarrow 1+2^{3}+3^{3}+...+n^{3}=1010^{2}.2021^{2}$
$\Leftrightarrow \left( 1+2+...+n \right)^{2}=1010^{2}.2021^{2}$
$\Leftrightarrow \left[ \frac{n\left( n+1 \right)}{2} \right]^{2}=1010^{2}.2021^{2}$
$\Leftrightarrow \frac{n\left( n+1 \right)}{2}=1010.2021$
$\Leftrightarrow n^{2}+n-2020.2021=0$
$\Leftrightarrow \left[ \begin{array}{l} n=2020 \\ n=-2021\,\,\left( \ell \right) \end{array} \right.$