The value of `(1)/(log_(2)n)+(1)/(log_(3)n)+....+(1)/(log_(43)n)` is …..

Answer

To solve the expression \( \frac{1}{\log_2 n} + \frac{1}{\log_3 n} + \ldots + \frac{1}{\log_{43} n} \), we can use the change of base formula for logarithms. The change of base formula states that:

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

Using this property, we can rewrite each term in the sum:

\[ \frac{1}{\log_k n} = \log_n k \]

Thus, we can rewrite the entire expression as:

\[ S = \log_n 2 + \log_n 3 + \ldots + \log_n 43 \]

Now, we can use the property of logarithms that states:

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

Applying this property, we can combine all the logarithmic terms:

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

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

\[ S = \log_n (43!) \]

This is our final answer.

Thus, the value of the expression \( \frac{1}{\log_2 n} + \frac{1}{\log_3 n} + \ldots + \frac{1}{\log_{43} n} \) is:

\[ \log_n (43!) \]