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

Options

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

Answer

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 use 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 this property to rewrite each term in the series.

### Step 1: Rewrite each term using the change of base formula

We can rewrite each term as follows:

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

Thus, the entire sum becomes:

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

### Step 2: Factor out \(\frac{1}{\log N}\)

Since \(\frac{1}{\log N}\) is a common factor in all terms, we can factor it out:

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

### Step 3: Simplify the sum of logarithms

The sum \(\sum_{k=2}^{100} \log k\) can be simplified using the property of logarithms that states \(\log a + \log b = \log(ab)\). Therefore, we have:

\[ \sum_{k=2}^{100} \log k = \log(2 \cdot 3 \cdot 4 \cdots \cdot 100). \]

This product can be expressed as \(100!\) (100 factorial):

\[ \sum_{k=2}^{100} \log k = \log(100!). \]

### Step 4: Combine the results

Now substituting back, we get:

\[ \frac{1}{\log N} \sum_{k=2}^{100} \log k = \frac{\log(100!)}{\log N}. \]

### Final Result

Thus, the final expression for the original series is:

\[ \frac{\log(100!)}{\log N}. \]