The value of `1/(log_100n)+1/(log_99n)+1/(log_98n)+...+1/(log_2n)` is :

Options

A. 1
B. `1/(log_1001n)`
C. `1/(log_n991)`
D. `1/(log_n1001)`

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 follow these steps:

### Step 1: Use the Change of Base Formula
We start by applying the change of base formula for logarithms, which states that:
\[ \log_a b = \frac{\log_c b}{\log_c a} \]
Using this property, we can rewrite each term in the sum:
\[ \frac{1}{\log_{k} n} = \frac{\log n}{\log k} \]
for \( k = 100, 99, 98, \ldots, 2 \).

### Step 2: Rewrite the Sum
Now we can rewrite the entire sum:
\[ \frac{1}{\log_{100} n} + \frac{1}{\log_{99} n} + \frac{1}{\log_{98} n} + \ldots + \frac{1}{\log_{2} n} = \log n \left( \frac{1}{\log 100} + \frac{1}{\log 99} + \frac{1}{\log 98} + \ldots + \frac{1}{\log 2} \right) \]

### Step 3: Simplify the Inner Sum
Let’s denote the inner sum as \( S \):
\[ S = \frac{1}{\log 100} + \frac{1}{\log 99} + \frac{1}{\log 98} + \ldots + \frac{1}{\log 2} \]
This sum can be expressed in terms of logarithms.

### Step 4: Use the Property of Logarithms
We can use the property of logarithms that states:
\[ \log a + \log b + \log c + \ldots = \log (a \cdot b \cdot c \cdots) \]
Thus, we can express \( S \) in terms of a single logarithm:
\[ S = \log \left( 100 \times 99 \times 98 \times \ldots \times 2 \right) \]
This product can be recognized as \( 100! \) (100 factorial).

### Step 5: Final Expression
Now, substituting back into our expression, we have:
\[ \frac{1}{\log_{100} n} + \frac{1}{\log_{99} n} + \frac{1}{\log_{98} n} + \ldots + \frac{1}{\log_{2} n} = \log n \cdot S = \log n \cdot \frac{1}{\log (100!)} \]
Thus, the final result is:
\[ \frac{\log n}{\log (100!)} \]

### Summary
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:
\[ \frac{\log n}{\log (100!)} \]