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प्रश्न
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
पर्याय
\[\frac{n (n + 1)}{2}\]
\[\frac{n (n + 1) (2n + 1)}{12}\]
\[\frac{n (n + 1)}{4}\]
none of these
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उत्तर
\[\frac{n (n + 1)}{4}\]
Let
\[S_n = \frac{1}{\log_2 4} + \frac{1}{\log_4 4} + \frac{1}{\log_8 4} + . . . + \frac{1}{\log_2^n 4}\]
\[\Rightarrow S_n = \frac{\log2}{\log4} + \frac{\log4}{\log4} + \frac{\log8}{\log4} + . . . + \frac{\log 2^n}{\log4}\]
\[ \Rightarrow S_n = \frac{\log2}{\log4} + \frac{\log 2^2}{\log4} + \frac{\log 2^3}{\log4} + . . . + \frac{\log 2^n}{\log4} \]
\[ \Rightarrow S_n = \frac{\log2}{\log4} + \frac{2 \log2}{\log4} + \frac{3 \log2}{\log4} + . . . + \frac{n \log2}{\log4} \]
\[ \Rightarrow S_n = \frac{\log2}{\log4}\left( 1 + 2 + 3 + . . . + n \right)\]
\[ \Rightarrow S_n = \frac{\log 4^\frac{1}{2}}{\log4}\left( 1 + 2 + 3 + . . . + n \right)\]
\[ \Rightarrow S_n = \frac{\frac{1}{2}\log4}{\log4}\left( 1 + 2 + 3 + . . . + n \right)\]
\[ \Rightarrow S_n = \frac{1}{2}\left( 1 + 2 + 3 + . . . + n \right)\]
\[ \Rightarrow S_n = \frac{n\left( n + 1 \right)}{4}\]
