# The Sum of the Series 1 Log 2 4 + 1 Log 4 4 + 1 Log 8 4 + . . . . + 1 Log N 2 4 is - Mathematics

MCQ

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

#### Solution

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

Let

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

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

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

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

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

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

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

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

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

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 21 Some special series
Q 2 | Page 19