Question

The sum $\displaystyle\sum_{ k =1}^{99} \log \sqrt[3]{\frac{ k }{ k +1}}$ is equal to (Assume base of logarithm is 10.)

• A. 2
• B. -2
• C. $\frac{-2}{3}$
• D. $\frac{2}{3}$

Answer 1

Answer: (C)

$S =\frac{1}{3}\left[\log \frac{1}{2}+\log \frac{2}{3}+\log \frac{3}{4}+\ldots \ldots+\log \frac{99}{100}\right]=\frac{1}{3}\left[\log \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \ldots \ldots \ldots \frac{99}{100}\right]$
$S =\frac{1}{3} \log _{10} \frac{1}{100}=\frac{-2}{3}$.