Question

${\sum }_{n=1}^n\frac{1}{lo{g}_{2^n}\left(a\right)}=$

• A. $\frac{n(n+1)}{2}lo{g}_{a}2$
• B. $\frac{n(n+1)}{2}lo{g}_{2}a$
• C. $\frac{(n+1{)}^2{n}^2}{4}lo{g}_{2}a$
• D. $Noneofthese$

Answer 1

The correct option is A $\frac{n(n+1)}{2}lo{g}_{a}2$

${\sum }_{n=1}^n\frac{1}{lo{g}_{2^n}\left(a\right)}$
$={\sum }_{n=1}^{n}lo{g}_a{2}^n$
$={\sum }_{n=1}^{n}nlo{g}_{a}2.$
$=lo{g}_{a}2.{\sum }_{n=1}^{n}n$
$=lo{g}_{a}2.\frac{n(n+1)}{2}$
$=\frac{n(n+1)}{2}lo{g}_{a}2$