The sum of the series $\frac{1}{\log _2(4)}+\frac{1}{\log _4(4)}+\frac{1}{\log _8(4)}+\ldots \ldots

Question

The sum of the series $\frac{1}{\log _2(4)}+\frac{1}{\log _4(4)}+\frac{1}{\log _8(4)}+\ldots \ldots \ldots \ldots \ldots \ldots . .+\frac{1}{\log _{2^n}(4)}$ is

Professor PC Thomas Classes · Answer

From given expression we get $$ \begin{aligned} & \log _4(2)+\log _4(4)+\log _4(8)+\ldots \ldots \ldots \ldots .+\log _4(2 n) \ & \log _{2^2}\left(2^1 	imes 2^2 \ldots \ldots \ldots \ldots 2^night) \ & =\frac{1+2+3+\ldots \ldots \ldots \ldots \ldots . n}{2}=\frac{n(n+1)}{4}\left(\because \log _2^2=1ight) \end{aligned} $$