What is the value of `(1)/(log_(2)n)+(1)/(log_(3)n)+ . . .+(1)/(log_(40)n)` ?

Options

A. `(1)/(log_((40!))n)`
B. `log_((40!)^(n))`
C. 1
D. None of these

Answer

To solve the problem, we need to evaluate the expression:

\[ S = \frac{1}{\log_2 n} + \frac{1}{\log_3 n} + \frac{1}{\log_4 n} + \ldots + \frac{1}{\log_{40} n} \]

### Step 1: Apply the Change of Base Formula
Using the change of base formula for logarithms, we can express \(\log_b a\) as \(\frac{\log_k a}{\log_k b}\) for any base \(k\). Here, we will use base \(n\):

\[ \log_b n = \frac{\log_n n}{\log_n b} = \frac{1}{\log_n b} \]

Thus, we can rewrite each term in the sum:

\[ \frac{1}{\log_b n} = \log_n b \]

### Step 2: Rewrite the Sum
Now, we can rewrite the entire sum \(S\):

\[ S = \log_n 2 + \log_n 3 + \log_n 4 + \ldots + \log_n 40 \]

### Step 3: Use the Property of Logarithms
Using the property of logarithms that states \(\log_a b + \log_a c = \log_a (bc)\), we can combine the logarithms:

\[ S = \log_n (2 \cdot 3 \cdot 4 \cdots 40) \]

### Step 4: Recognize the Product
The product \(2 \cdot 3 \cdot 4 \cdots 40\) is equal to \(40!\) (40 factorial). Therefore, we can express \(S\) as:

\[ S = \log_n (40!) \]

### Final Step: Conclusion
Thus, the value of the original expression is:

\[ S = \log_n (40!) \]