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

Options

• \[\frac{n (n + 1)}{2}\]

• \[\frac{n (n + 1) (2n + 1)}{12}\]

• \[\frac{n (n + 1)}{4}\]

• none of these

Answer 1

\[\frac{n (n + 1)}{4}\]

Let

\[S_n = \frac{1}{\log_2 4} + \frac{1}{\log_4 4} + \frac{1}{\log_8 4} + . . . + \frac{1}{\log_2^n 4}\]

\[\Rightarrow S_n = \frac{\log2}{\log4} + \frac{\log4}{\log4} + \frac{\log8}{\log4} + . . . + \frac{\log 2^n}{\log4}\]

\[ \Rightarrow S_n = \frac{\log2}{\log4} + \frac{\log 2^2}{\log4} + \frac{\log 2^3}{\log4} + . . . + \frac{\log 2^n}{\log4} \]

\[ \Rightarrow S_n = \frac{\log2}{\log4} + \frac{2 \log2}{\log4} + \frac{3 \log2}{\log4} + . . . + \frac{n \log2}{\log4} \]

\[ \Rightarrow S_n = \frac{\log2}{\log4}\left( 1 + 2 + 3 + . . . + n \right)\]

\[ \Rightarrow S_n = \frac{\log 4^\frac{1}{2}}{\log4}\left( 1 + 2 + 3 + . . . + n \right)\]

\[ \Rightarrow S_n = \frac{\frac{1}{2}\log4}{\log4}\left( 1 + 2 + 3 + . . . + n \right)\]

\[ \Rightarrow S_n = \frac{1}{2}\left( 1 + 2 + 3 + . . . + n \right)\]

\[ \Rightarrow S_n = \frac{n\left( n + 1 \right)}{4}\]