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Câu hỏi: Tìm $n\in {{\mathbb{N}}^{*}}$ biết $\dfrac{1}{{{\log }_{2}}x}+\dfrac{1}{{{\log }_{{{2}^{2}}}}x}+\dfrac{1}{{{\log }_{{{2}^{3}}}}x}+...+\dfrac{1}{{{\log }_{{{2}^{n}}}}x}=\dfrac{465}{{{\log }_{2}}x}$ luôn đúng với mọi $x>0,x\ne 1.$
A. $n=31.$
B. $n\in \varnothing $.
C. $n=30.$
D. $n=-31.$
A. $n=31.$
B. $n\in \varnothing $.
C. $n=30.$
D. $n=-31.$
Ta có $\dfrac{1}{{{\log }_{2}}x}+\dfrac{1}{{{\log }_{{{2}^{2}}}}x}+\dfrac{1}{{{\log }_{{{2}^{3}}}}x}+...+\dfrac{1}{{{\log }_{{{2}^{n}}}}x}={{\log }_{x}}2+{{\log }_{x}}{{2}^{2}}+{{\log }_{x}}{{2}^{3}}+...+lo{{g}_{x}}{{2}^{n}}$
$={{\log }_{x}}\left( {{2.2}^{2}}{{.2}^{3}}{{...2}^{n}} \right)=465{{\log }_{x}}2={{\log }_{x}}{{2}^{465}}$
$\Leftrightarrow 1+2+3+...+n=465\Leftrightarrow \dfrac{n}{2}\left( n+1 \right)=465$
$\Leftrightarrow {{n}^{2}}+n-930=0\Leftrightarrow \left[ \begin{aligned}
& n=30 \\
& n=-31 \\
\end{aligned} \right.\Rightarrow n=30.$
$={{\log }_{x}}\left( {{2.2}^{2}}{{.2}^{3}}{{...2}^{n}} \right)=465{{\log }_{x}}2={{\log }_{x}}{{2}^{465}}$
$\Leftrightarrow 1+2+3+...+n=465\Leftrightarrow \dfrac{n}{2}\left( n+1 \right)=465$
$\Leftrightarrow {{n}^{2}}+n-930=0\Leftrightarrow \left[ \begin{aligned}
& n=30 \\
& n=-31 \\
\end{aligned} \right.\Rightarrow n=30.$
Đáp án C.