Tìm n biết 1/log2 x + 1/log2^2 x + 1/log2^3 x + ...+ 1/log2^n x = 465/log2 x luôn

* Hướng dẫn giải

Đáp án C

Ta có

$$\begin{gathered} \frac{1}{{\log }_{2}x}+\frac{1}{{\log }_{2^2}x}+\frac{1}{{\log }_{2^3}x}+\mathrm{...}+\frac{1}{{\log }_{2^n}x} \\ ={\log }_{x}2+{\log }_x{2}^2+{\log }_x{2}^3+\mathrm{...}+{\log }_x{2}^n \end{gathered}$$

$$\begin{gathered} \begin{array}{l}={\log }_x\left({2.2}^2{.2}^3{\mathrm{....2}}^n\right)=465{\log }_{x}2={\log }_x{2}^{465}\Rightarrow {2.2}^2{.2}^3{\mathrm{...2}}^n=2^{465}.\\ \iff 1+2+3+\mathrm{...}+n=465\iff \frac{n\left(n+1\right)}{2}=465\iff n^2+n-930=0\end{array} \\ \iff \left[\begin{array}{l}n=30\\ n=-31\end{array}\right.\Rightarrow n=30. \end{gathered}$$