`N=n!`, where `ngt2`. Find the value of `(log_(2)N)^(-1)+(log_(3)N)^(-1)+(log_(4)N)^(-1)+. . . . . (log_(n)N)^(-1)`.

Options

A. (A)0
B. (B)1
C. (C)10
D. (D)N

Answer

To solve the problem, we need to find the value of the expression:

\[ \frac{1}{\log_2 N} + \frac{1}{\log_3 N} + \frac{1}{\log_4 N} + \ldots + \frac{1}{\log_n N} \]

where \( N = n! \) and \( n > 2 \).

### Step 1: Rewrite the expression using the change of base formula
Using the property of logarithms that states:

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

we can rewrite each term in the expression:

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

Thus, the expression becomes:

\[ \log_N 2 + \log_N 3 + \log_N 4 + \ldots + \log_N n \]

### Step 2: Combine the logarithms
Using the property of logarithms that states:

\[ \log_a b + \log_a c = \log_a (bc) \]

we can combine the logarithms:

\[ \log_N (2 \times 3 \times 4 \times \ldots \times n) \]

### Step 3: Recognize the product as \( n! \)
The product \( 2 \times 3 \times 4 \times \ldots \times n \) is equal to \( n! \). Therefore, we can rewrite the expression as:

\[ \log_N (n!) \]

### Step 4: Substitute \( N \) with \( n! \)
Since \( N = n! \), we have:

\[ \log_{n!} (n!) \]

### Step 5: Evaluate the logarithm
Using the property of logarithms that states:

\[ \log_a a = 1 \]

we find:

\[ \log_{n!} (n!) = 1 \]

### Final Answer
Thus, the value of the original expression is:

\[ \boxed{1} \]