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Câu hỏi: Tìm số nguyên dương $n$ sao cho
${{\log }_{2018}}2019+{{2}^{2}}{{\log }_{\sqrt{2018}}}2019+{{3}^{2}}{{\log }_{\sqrt[3]{2018}}}2019+...+{{n}^{2}}{{\log }_{\sqrt[n]{2018}}}2019={{1010}^{2}}{{.2021}^{2}}{{\log }_{2018}}2019$
A. $n=2021$
B. $n=2019$
C. $n=2020$
D. $n=2018$
${{\log }_{2018}}2019+{{2}^{2}}{{\log }_{\sqrt{2018}}}2019+{{3}^{2}}{{\log }_{\sqrt[3]{2018}}}2019+...+{{n}^{2}}{{\log }_{\sqrt[n]{2018}}}2019={{1010}^{2}}{{.2021}^{2}}{{\log }_{2018}}2019$
A. $n=2021$
B. $n=2019$
C. $n=2020$
D. $n=2018$
Ta có: ${{k}^{2}}{{\log }_{\sqrt[k]{2018}}}2019={{k}^{2}}{{\log }_{{{2018}^{\dfrac{1}{k}}}}}2019={{k}^{3}}{{\log }_{2018}}2019$
Theo đề:
${{\log }_{2018}}2019+{{2}^{2}}{{\log }_{\sqrt{2018}}}2019+{{3}^{2}}{{\log }_{\sqrt[3]{2018}}}2019+...+{{n}^{2}}{{\log }_{\sqrt[n]{2018}}}2019={{1010}^{2}}{{.2021}^{2}}{{\log }_{2018}}2019$
$\Leftrightarrow {{1}^{3}}.{{\log }_{2018}}2019+{{2}^{3}}{{\log }_{2018}}2019+{{3}^{3}}{{\log }_{2018}}2019+...+{{n}^{3}}{{\log }_{2018}}2019={{1010}^{2}}{{.2021}^{2}}{{\log }_{2018}}2019$
$\Leftrightarrow \left( {{1}^{3}}+{{2}^{3}}+{{3}^{3}}+...+{{n}^{3}} \right){{\log }_{2018}}2019={{1010}^{2}}{{.2021}^{2}}{{\log }_{2018}}2019$
$\Leftrightarrow {{1}^{3}}+{{2}^{3}}+{{3}^{3}}+...+{{n}^{3}}={{1010}^{2}}{{.2021}^{2}} (1)$
Theo chương trình lớp 11 ta có: ${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+...+{{n}^{3}}={{\left[ \dfrac{n\left( n+1 \right)}{2} \right]}^{2}}$
$(1) \Leftrightarrow \dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}={{1010}^{2}}{{.2021}^{2}}\Leftrightarrow {{n}^{2}}{{\left( n+1 \right)}^{2}}={{4.1010}^{2}}{{.2021}^{2}}\Leftrightarrow {{n}^{2}}{{\left( n+1 \right)}^{2}}={{2020}^{2}}{{.2021}^{2}}$
$\Leftrightarrow n=2020$
Theo đề:
${{\log }_{2018}}2019+{{2}^{2}}{{\log }_{\sqrt{2018}}}2019+{{3}^{2}}{{\log }_{\sqrt[3]{2018}}}2019+...+{{n}^{2}}{{\log }_{\sqrt[n]{2018}}}2019={{1010}^{2}}{{.2021}^{2}}{{\log }_{2018}}2019$
$\Leftrightarrow {{1}^{3}}.{{\log }_{2018}}2019+{{2}^{3}}{{\log }_{2018}}2019+{{3}^{3}}{{\log }_{2018}}2019+...+{{n}^{3}}{{\log }_{2018}}2019={{1010}^{2}}{{.2021}^{2}}{{\log }_{2018}}2019$
$\Leftrightarrow \left( {{1}^{3}}+{{2}^{3}}+{{3}^{3}}+...+{{n}^{3}} \right){{\log }_{2018}}2019={{1010}^{2}}{{.2021}^{2}}{{\log }_{2018}}2019$
$\Leftrightarrow {{1}^{3}}+{{2}^{3}}+{{3}^{3}}+...+{{n}^{3}}={{1010}^{2}}{{.2021}^{2}} (1)$
Theo chương trình lớp 11 ta có: ${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+...+{{n}^{3}}={{\left[ \dfrac{n\left( n+1 \right)}{2} \right]}^{2}}$
$(1) \Leftrightarrow \dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}={{1010}^{2}}{{.2021}^{2}}\Leftrightarrow {{n}^{2}}{{\left( n+1 \right)}^{2}}={{4.1010}^{2}}{{.2021}^{2}}\Leftrightarrow {{n}^{2}}{{\left( n+1 \right)}^{2}}={{2020}^{2}}{{.2021}^{2}}$
$\Leftrightarrow n=2020$
Đáp án C.