Question

$\frac{1}{{\log }_{2}n}+\frac{1}{{\log }_{3}n}+\frac{1}{{\log }_{4}n}+...\frac{1}{{\log }_{43}n}=$

• A. $1$
• B. ${\log }_{43!}n$
• C. ${\log }_{n}43!$
• D. $none$

Answer 1

Answer: (C)

${\log }_{n}43!$

$\frac{1}{lo{g}_{2}n}+\frac{1}{lo{g}_{3}n}+\frac{1}{lo{g}_{4}n}+..........+\frac{1}{lo{g}_{43}n}$

$\frac{log2}{logn}+\frac{log3}{logn}+\frac{log4}{logn}+......+\frac{log43}{logn}$

$\frac{log2+log3+log4+.......+log43}{logn}$

$\frac{log\left(\mathrm{2.3.4.......43}\right)}{logn}$

$\frac{log43!}{logn}=lo{g}_n(43!)$

Option 3 is correct answer