Question

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

• A. $\frac{n(n+1)}{2}$
• B. $\frac{n(n+1)(2n+1)}{12}$
• C. $\frac{n(n+1)}{4}$
• D. none of these

Answer 1

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

$Let{S}_n=\frac{1}{lo{g}_{2}4}+\frac{1}{lo{g}_{4}4}+\frac{1}{lo{g}_{8}4}+...+\frac{1}{lo{g}_{2^n}4}\Rightarrow S_n=\frac{log2}{log4}+\frac{log4}{log4}+\frac{log8}{log4}+...+\frac{log{2}^2}{log4}\Rightarrow S_n=\frac{log2}{log4}+\frac{log{2}^2}{log4}+\frac{log{2}^3}{log4}+...+\frac{log{2}^n}{log4}\Rightarrow S_n=\frac{log2}{log4}+\frac{2log2}{log4}+\frac{3log2}{log4}+...+\frac{nlog2}{log4}\Rightarrow S_n=\frac{log2}{log4}+(1+2+3+...+n)\Rightarrow S_n=\frac{log{4}^{\frac{1}{2}}}{log4}(1+2+3+....+n)\Rightarrow S_n=\frac{\frac{1}{2}}{log4}(1+2+3+....+n)\Rightarrow S_n=\frac{1}{2}(1+2+3+....+n)\Rightarrow S_n=\frac{\left(n\right(n+1\left)\right)}{4}$