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Câu hỏi: Tìm số nguyên dương n thõa mãn:
${{\log }_{a}}2020+\dfrac{1}{{{2}^{2}}}{{\log }_{\sqrt{a}}}2020+\dfrac{1}{{{2}^{4}}}{{\log }_{\sqrt[4]{a}}}2020+\dfrac{1}{{{2}^{6}}}{{\log }_{\sqrt[8]{a}}}2020...+\dfrac{1}{{{2}^{2n}}}{{\log }_{\sqrt[2n]{a}}}2020={{\log }_{a}}{{2020}^{2}}-\dfrac{{{\log }_{a}}2020}{{{2}^{2021}}},$
với $0<a\ne 1$
A. n = 2019
B. n = 2020
C. n = 2021
D. n = 2022
${{\log }_{a}}2020+\dfrac{1}{{{2}^{2}}}{{\log }_{\sqrt{a}}}2020+\dfrac{1}{{{2}^{4}}}{{\log }_{\sqrt[4]{a}}}2020+\dfrac{1}{{{2}^{6}}}{{\log }_{\sqrt[8]{a}}}2020...+\dfrac{1}{{{2}^{2n}}}{{\log }_{\sqrt[2n]{a}}}2020={{\log }_{a}}{{2020}^{2}}-\dfrac{{{\log }_{a}}2020}{{{2}^{2021}}},$
với $0<a\ne 1$
A. n = 2019
B. n = 2020
C. n = 2021
D. n = 2022
Gọi vế trái và vế phải của hệ thức đề bài cho lần lượt là A và B.
Ta có $\dfrac{1}{{{2}^{2n}}}{{\log }_{\sqrt[{{2}^{n}}]{a}}}2020=\dfrac{1}{{{2}^{2n}}}{{\log }_{a}}{{2020}^{{{2}^{n}}}}=\dfrac{{{2}^{n}}}{{{2}^{2n}}}.{{\log }_{a}}2020$
Do đó
$\begin{aligned}
& A={{\log }_{a}}2020+\dfrac{1}{4}{{\log }_{\sqrt{a}}}2020+...+\dfrac{1}{{{2}^{2n}}}.{{\log }_{\sqrt[{{2}^{n}}]{a}}}2020 \\
& =\left( 1+\dfrac{{{2}^{1}}}{4}+\dfrac{{{2}^{2}}}{8}+\dfrac{{{2}^{3}}}{16}+...+\dfrac{{{2}^{n}}}{{{2}^{2n}}} \right){{\log }_{a}}2020 \\
& =\left( 1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{{{2}^{n}}} \right){{\log }_{a}}2020 \\
& =\left( \dfrac{1-\dfrac{1}{{{2}^{n+1}}}}{1-\dfrac{1}{2}} \right){{\log }_{a}}2020 \\
& =\left( 2-\dfrac{1}{{{2}^{n}}} \right)\log {}_{a}2020 \\
\end{aligned}$
$={{\log }_{a}}{{2020}^{2}}-\dfrac{1}{{{2}^{n}}}{{\log }_{a}}2020$ mà $A={{\log }_{a}}{{2020}^{2}}-\dfrac{{{\log }_{a}}2020}{{{2}^{2021}}}$
Suy ra n = 2021
Ta có $\dfrac{1}{{{2}^{2n}}}{{\log }_{\sqrt[{{2}^{n}}]{a}}}2020=\dfrac{1}{{{2}^{2n}}}{{\log }_{a}}{{2020}^{{{2}^{n}}}}=\dfrac{{{2}^{n}}}{{{2}^{2n}}}.{{\log }_{a}}2020$
Do đó
$\begin{aligned}
& A={{\log }_{a}}2020+\dfrac{1}{4}{{\log }_{\sqrt{a}}}2020+...+\dfrac{1}{{{2}^{2n}}}.{{\log }_{\sqrt[{{2}^{n}}]{a}}}2020 \\
& =\left( 1+\dfrac{{{2}^{1}}}{4}+\dfrac{{{2}^{2}}}{8}+\dfrac{{{2}^{3}}}{16}+...+\dfrac{{{2}^{n}}}{{{2}^{2n}}} \right){{\log }_{a}}2020 \\
& =\left( 1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{{{2}^{n}}} \right){{\log }_{a}}2020 \\
& =\left( \dfrac{1-\dfrac{1}{{{2}^{n+1}}}}{1-\dfrac{1}{2}} \right){{\log }_{a}}2020 \\
& =\left( 2-\dfrac{1}{{{2}^{n}}} \right)\log {}_{a}2020 \\
\end{aligned}$
$={{\log }_{a}}{{2020}^{2}}-\dfrac{1}{{{2}^{n}}}{{\log }_{a}}2020$ mà $A={{\log }_{a}}{{2020}^{2}}-\dfrac{{{\log }_{a}}2020}{{{2}^{2021}}}$
Suy ra n = 2021
Đáp án C.