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 equal to

• A. $\frac{n(n+1)}{2}$
• B. $\frac{n(n+1)(2n+1)}{12}$
• C. $\frac{1}{n(n+1)}$
• D. $\frac{n(n+1)}{4}$

Answer 1

Answer: (D)

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

Let $S=\frac{1}{{\log }_{2}4}+\frac{1}{{\log }_{4}4}+\frac{1}{{\log }_{8}4}+...+\frac{1}{{\log }_{2^n}4}$
$\Rightarrow S=\frac{{\log }_{2}2+{\log }_2{2}^2+{\log }_2{2}^3+....+{\log }_2{2}^n}{2}$
$\Rightarrow S=\frac{1}{2}{\sum }_{r=1}^n{\log }_2{2}^r=\frac{1}{2}{\sum }_{r=1}^{n}r$
$\Rightarrow S=\frac{n(n+1)}{4}$
Ans: D