Let $n = 2006!$. Then $\frac{1}{\log_{2}n} + \frac{1}{\log_{3}n} + ... + \frac{1}{\log_{2006}n}$ is equal to

Question

Dr. Abubacker · Accepted Answer

(d)
Given, $n = 2006!$
$	herefore\ \frac{1}{\log_{2}n} + \frac{1}{\log_{3}n} + ... + \frac{1}{\log_{2006}n}$
$= \log_{n}{2 + \log_{n}{3 + ... + \log_{n}2006}}$
$= \log_{n}{(2.3.4.\ \ldots.2006)}$
$= \log_{n}{(2006!) = \log_{n}{n = 1}}$

Q. Let $n = 2006!$. Then $\frac{1}{\log_{2}n} + \frac{1}{\log_{3}n} + ... + \frac{1}{\log_{2006}n}$ is equal to

Explanation

Reviewed by Dr. Abubacker

Want more practice?