Question

If $N=n!$, where n is a'fixed integer $>2$, then $\frac{1}{{\log }_{2}N}+\frac{1}{{\log }_{3}N}+...+\frac{1}{{\log }_{n}N}=$

• A. -1
• B. 0
• C. 1
• D. 4

Answer 1

Answer: (C)

$\frac{1}{{\log }_{2}N}+\frac{1}{{\log }_{3}N}+...+\frac{1}{{\log }_{n}N}$
$=\frac{\log 2}{\log N}+\frac{\log 3}{\log N}+.....+\frac{\log n}{\log N}$
$=\frac{1}{\log N}\left(\log 2+\log 3+...+\log n\right)$
$=\frac{1}{\log N}\log \left(\mathrm{2.3.4.....}n\right)$
$=\frac{1}{\log n!}\log \left(\mathrm{1.2.3.4....}n\right)\left(\because N=n!\right)$
$=\frac{\log \left(n!\right)}{\log \left(n!\right)}=1$