What is `(1)/(log_(2)N)+(1)/(log_(3)N)+(1)/(log_(4)N)+....+(1)/(log_(100)N) " equal to "(Nne1)`?

Options

A. `(1)/(log_(100!)N)`
B. `(1)/(log_(99!)N)`
C. `(99)/(log_(100!)N)`
D. `(99)/(log_(99!)N)`

Explanation

To solve the expression

\[ \frac{1}{\log_2 N} + \frac{1}{\log_3 N} + \frac{1}{\log_4 N} + \ldots + \frac{1}{\log_{100} N}, \]

we can start by rewriting each term using the change of base formula for logarithms. The change of base formula states that

\[ \log_a b = \frac{\log_c b}{\log_c a} \]

for any positive base \(c\). We will use base \(N\) for our calculations.

### Step 1: Rewrite each term

Using the change of base formula, we can rewrite each term as follows:

\[ \frac{1}{\log_k N} = \frac{\log N}{\log k} \]

Thus, the entire sum becomes:

\[ \sum_{k=2}^{100} \frac{1}{\log_k N} = \sum_{k=2}^{100} \frac{\log N}{\log k} \]

### Step 2: Factor out \(\log N\)

Since \(\log N\) is a common factor in each term of the sum, we can factor it out:

\[ \sum_{k=2}^{100} \frac{\log N}{\log k} = \log N \sum_{k=2}^{100} \frac{1}{\log k} \]

### Step 3: Calculate the sum \(\sum_{k=2}^{100} \frac{1}{\log k}\)

Now, we need to evaluate the sum \(\sum_{k=2}^{100} \frac{1}{\log k}\). This sum does not have a simple closed form, but we can denote it as \(S\):

\[ S = \sum_{k=2}^{100} \frac{1}{\log k} \]

### Step 4: Final expression

Now, we can express our original sum in terms of \(S\):

\[ \frac{1}{\log_2 N} + \frac{1}{\log_3 N} + \ldots + \frac{1}{\log_{100} N} = \log N \cdot S \]

### Step 5: Relate to \(N^{\frac{1}{\log 100!}}\)

We can relate this to the expression \(N^{\frac{1}{\log 100!}}\). We know that:

\[ \log 100! = \sum_{k=1}^{100} \log k \]

Thus, we can express the final result as:

\[ \frac{\log N \cdot S}{\log 100!} = N^{\frac{1}{\log 100!}} \]

### Conclusion

Therefore, we conclude that:

\[ \frac{1}{\log_2 N} + \frac{1}{\log_3 N} + \ldots + \frac{1}{\log_{100} N} = \frac{\log N}{\log 100!} \]