The value of `1/(log_(100)n) + 1/(log_(99)n) + 1/(log_(98)n) +`……..`+1/(log_(2)n)` is,

Options

A. 1
B. `1/(log_(1001)n)`
C. `1/(log_(n)991)`
D. `1/(log_(n)1001)`

Answer

To solve the problem \( \frac{1}{\log_{100} n} + \frac{1}{\log_{99} n} + \frac{1}{\log_{98} n} + \ldots + \frac{1}{\log_{2} n} \), we can use the change of base formula for logarithms. The change of base formula states that:

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

This means we can rewrite each term in the series.

### Step-by-step Solution:

1. **Rewrite Each Term Using Change of Base Formula**:
\[ \frac{1}{\log_{k} n} = \log_{n} k \]
Therefore, we can rewrite the entire sum as:
\[ \log_{n} 100 + \log_{n} 99 + \log_{n} 98 + \ldots + \log_{n} 2 \]

2. **Combine the Logarithms**:
Using the property of logarithms that states \( \log_a b + \log_a c = \log_a (b \cdot c) \), we can combine the logarithmic terms:
\[ \log_{n} (100 \cdot 99 \cdot 98 \cdots 2) \]

3. **Recognize the Product as a Factorial**:
The product \( 100 \cdot 99 \cdot 98 \cdots 2 \) can be expressed as \( 100! \) (100 factorial):
\[ \log_{n} (100!) \]

4. **Final Expression**:
Thus, the entire expression simplifies to:
\[ \log_{n} (100!) \]

### Final Answer:
The value of \( \frac{1}{\log_{100} n} + \frac{1}{\log_{99} n} + \frac{1}{\log_{98} n} + \ldots + \frac{1}{\log_{2} n} \) is:
\[ \log_{n} (100!) \]