Question

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

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

Answer 1

Answer: (A)

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

Series is $\frac{1}{{\log }_{2}4}+\frac{1}{{\log }_{4}4}+\frac{1}{{\log }_{8}4}+........+\frac{1}{{\log }_{2^n}4}$

$=\frac{1}{{\log }_2{2}^2}+\frac{1}{{\log }_{2^2}{2}^2}+\frac{1}{{\log }_{2^3}{2}^2}+.......+\frac{1}{{\log }_{2^n}{2}^2}=\frac{1}{2\log 2^2}+\frac{1}{2\times \frac{1}{2}\log 2^2}+\frac{1}{2\times \frac{1}{3}\log 2^2}+...........+\frac{1}{2\times \frac{1}{n}\log 2^2}[\log a^a=1]=\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+.......+\frac{n}{2}=\frac{1}{2}[1+2+3+4+......+n]=\frac{1}{2}\frac{n(n+1)}{2}=\frac{n(n+1)}{4}$