The sum of the series $\frac{1}{\log _{2} 4}+\frac{1}{\log _{4} 4}+\frac{1}{\log _{8} 4}+\ldots+\frac{1}{\log _{2}{ }^{n} 4}$ is
(a) $\frac{n(n+1)}{2}$
(b) $\frac{n(n+1)(2 n+1)}{12}$
(c) $\frac{n(n+1)}{4}$
(d) none of these
(c) $\frac{n(n+1)}{4}$
Let $S_{n}=\frac{1}{\log _{2} 4}+\frac{1}{\log _{4} 4}+\frac{1}{\log _{8} 4}+\ldots+\frac{1}{\log _{2}{ }_{2} 4}$
$\Rightarrow S_{n}=\frac{\log 2}{\log 4}+\frac{\log 4}{\log 4}+\frac{\log 8}{\log 4}+\ldots+\frac{\log 2^{n}}{\log 4}$
$\Rightarrow S_{n}=\frac{\log 2}{\log 4}+\frac{\log 2^{2}}{\log 4}+\frac{\log 2^{3}}{\log 4}+\ldots+\frac{\log 2^{n}}{\log 4}$
$\Rightarrow S_{n}=\frac{\log 2}{\log 4}+\frac{2 \log 2}{\log 4}+\frac{3 \log 2}{\log 4}+\ldots+\frac{n \log 2}{\log 4}$
$\Rightarrow S_{n}=\frac{\log 2}{\log 4}(1+2+3+\ldots+n)$
$\Rightarrow S_{n}=\frac{\log 4 / 2}{\log 4}(1+2+3+\ldots+n)$
$\Rightarrow S_{n}=\frac{\frac{1}{\log 4}}{\log 4}(1+2+3+\ldots+n)$
$\Rightarrow S_{n}=\frac{1}{2}(1+2+3+\ldots+n)$
$\Rightarrow S_{n}=\frac{n(n+1)}{4}$
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