`(1)/("log"_(2)n) + (1)/("log"_(3)n) + (1)/("log"_(4)n) + … + (1)/("log"_(43)n)=`

Options

A. 1
B. `"log"_(43!)n`
C. `"log"_(n)43!`
D. none of these

Explanation

To solve the equation

\[ \frac{1}{\log_2 n} + \frac{1}{\log_3 n} + \frac{1}{\log_4 n} + \ldots + \frac{1}{\log_{43} n} = ? \]

we can use the property of logarithms that states:

\[ \log_a b = \frac{1}{\log_b a} \]

This means we can rewrite each term in the sum. Thus, we have:

\[ \frac{1}{\log_2 n} = \log_n 2, \quad \frac{1}{\log_3 n} = \log_n 3, \quad \frac{1}{\log_4 n} = \log_n 4, \quad \ldots, \quad \frac{1}{\log_{43} n} = \log_n 43 \]

Now, substituting these into our original equation gives us:

\[ \log_n 2 + \log_n 3 + \log_n 4 + \ldots + \log_n 43 \]

Using the property of logarithms that states:

\[ \log_a b + \log_a c = \log_a (bc) \]

we can combine all these logarithms into a single logarithm:

\[ \log_n (2 \cdot 3 \cdot 4 \cdots 43) \]

The product \(2 \cdot 3 \cdot 4 \cdots 43\) is the factorial of 43, denoted as \(43!\). Therefore, we can rewrite our equation as:

\[ \log_n (43!) \]

Thus, the final answer is:

\[ \log_n (43!) \]