What is 1 _2 N + 1 _3 N + 1 _4 N + .+ 1 _ 100 N equal to (N ≠ 1) ?

What is

\frac{1}{\log _2 \mathrm{ N}}+\frac{1}{\log _3 \mathrm{ N}}+\frac{1}{\log _4 \mathrm{ N}}+\ldots .+\frac{1}{\log _{100} \mathrm{ N}}

equal to

(\mathrm{N} \neq 1)

Options

A. $\frac{1}{\log _{100!} N}$

\begin{aligned} & \frac{1}{\log _2 \mathrm{ N}}+\frac{1}{\log _3 \mathrm{ N}}+\frac{1}{\log _4 \mathrm{ N}}+\ldots+\frac{1}{\log _{100} \mathrm{ N}} \\ & =\frac{\frac{1}{\log \mathrm{ N}}}{\log 2}+\frac{1}{\log \mathrm{ N}}+\frac{1}{\log 3}+\ldots+\frac{1}{\frac{\log \mathrm{ N}}{\log 4}} \\ & \quad\left(\because \log _{\mathrm{a}} \mathrm{b}=\frac{\log \mathrm{b}}{\log \mathrm{N}}\right) \\ & =\frac{\log 2}{\log \mathrm{ N}}+\frac{\log 3}{\log \mathrm{ N}}+\frac{\log 4}{\log \mathrm{ N}}+\ldots.+\frac{\log 100}{\log \mathrm{ N}} \\ & =\frac{\log 2+\log 3+\log 4+\ldots.+\log 100}{\log \mathrm{ N}} \\ & =\frac{\log (2.3.4 \ldots.100)}{\log \mathrm{N}}\end{aligned}

=\frac{\log 100!}{\log \mathrm{N}}=\frac{1}{\frac{\log \mathrm{ N}}{\log 100!}}=\frac{1}{\log _{100!} \mathrm{N}}$