Question

The sum of the series $\frac{1}{{\log }_{2}4}+\frac{1}{{\log }_{4}4}+\frac{1}{{\log }_{8}4}+....+\frac{1}{{{\log }_2}^{n}4}$ is
(a) $\frac{n(n+1)}{2}$

(b) $\frac{n(n+1)(2n+1)}{12}$

(c) $\frac{n(n+1)}{4}$

(d) none of these

Answer 1

(c) $\frac{n(n+1)}{4}$

Let $S_n=\frac{1}{{\log }_{2}4}+\frac{1}{{\log }_{4}4}+\frac{1}{{\log }_{8}4}+...+\frac{1}{{{\log }_2}^{n}4}$

$$\begin{gathered} \Rightarrow S_n=\frac{\log 2}{\log 4}+\frac{\log 4}{\log 4}+\frac{\log 8}{\log 4}+...+\frac{\log 2^n}{\log 4} \\ \Rightarrow S_n=\frac{\log 2}{\log 4}+\frac{\log 2^2}{\log 4}+\frac{\log 2^3}{\log 4}+...+\frac{\log 2^n}{\log 4} \\ \Rightarrow S_n=\frac{\log 2}{\log 4}+\frac{2\log 2}{\log 4}+\frac{3\log 2}{\log 4}+...+\frac{n\log 2}{\log 4} \\ \Rightarrow S_n=\frac{\log 2}{\log 4}\left(1+2+3+...+n\right) \\ \Rightarrow S_n=\frac{\log 4^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{\log 4}\left(1+2+3+...+n\right) \\ \Rightarrow S_n=\frac{{\displaystyle \frac{1}{2}}\log 4}{\log 4}\left(1+2+3+...+n\right) \\ \Rightarrow S_n=\frac{1}{2}\left(1+2+3+...+n\right) \\ \Rightarrow S_n=\frac{n\left(n+1\right)}{4} \end{gathered}$$