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

Answer 1

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

$\frac{1}{lo{g}_{2}4}+\frac{1}{lo{g}_{4}4}+\frac{1}{lo{g}_{3}4}+...\dots +\frac{1}{lo{g}_2^{n}4}$

$\Rightarrow \frac{lo{g}_{2}2}{lo{g}_{2}4}+\frac{lo{g}_{2}4}{lo{g}_{2}4}+\frac{lo{g}_{2}8}{lo{g}_{2}4}+...\dots +\frac{lo{g}_2{2}^n}{lo{g}_{2}4}$ $[lo{g}_{a}b=\frac{lo{g}_{c}b}{lo{g}_{c}a}]$

$\Rightarrow \frac{lo{g}_2{2}^1}{lo{g}_{2}4}+\frac{lo{g}_2{2}^2}{lo{g}_{2}4}+\frac{lo{g}_2{2}^3}{lo{g}_{2}4}+.....+\frac{lo{g}_2{2}^n}{lo{g}_{2}4}$

$\Rightarrow \frac{1}{lo{g}_{2}4}[1+2+3+...\dots +n]$ $[lo{g}_a{b}^n=nlo{g}_{a}b]$

$\Rightarrow \frac{1}{2}\left[\frac{n(n+1)}{2}\right]=\frac{1}{4}\left[n\right(n+1\left)\right]$.