Question

The sum of the series $\frac{1}{{\log }_{2}4}+\frac{1}{{\log }_{4}4}+\frac{1}{{\log }_{8}4}+...........+\frac{1}{{\log }_{2^n}4}$

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

Answer 1

The correct option is D $\frac{n(n+1)}{4}$

$\frac{1}{{\log }_{2}4}+\frac{1}{{\log }_{4}4}+\frac{1}{{\log }_{8}4}+....\frac{1}{{\log }_{2^n}4}$
$={\log }_{4}2+{\log }_{4}4+{\log }_{4}8+...+{\log }_4{2}^n$
$={\log }_4{4}^{1/2}+{\log }_{4}4+{\log }_4{4}^{3/2}+....{\log }_4{4}^{n/2}$
$=\frac{1}{2}+1+\frac{3}{2}+2+....\frac{n}{2}$
$=\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+....\frac{n}{2}$
$=\frac{1}{2}(1+2+3+...n)$
$=\frac{1}{2}\frac{n(n+1)}{2}$
$=\frac{n(n+1)}{4}$