Question

${\log }_{n}1+{\log }_n(1+\frac{1}{2})+{\log }_n(1+\frac{1}{3})+\dots \dots +{\log }_n(1+\frac{1}{n-1})=$

• A. $1-{\log }_{n}2$
• B. $1+{\log }_{n}2$
• C. ${\log }_2(n-1)$
• D. ${\log }_n(n+1)$

Answer 1

The correct option is A $1-{\log }_{n}2$

${\log }_{n}1+{\log }_n(1+\frac{1}{2})+{\log }_n(1+\frac{1}{3})+\dots \dots +{\log }_n(1+\frac{1}{n-1})$
$={\log }_n(1\cdot \frac{3}{2}\cdot \frac{4}{3}\cdots \frac{n}{n-1})={\text{log}}_n\left(\frac{n}{2}\right)=1-{\text{log}}_{n}2$