If `n=1983!`, then the value of expression `(1)/(log_(2)n)+(1)/(log_(3)n)+(1)/(log_(4)n.)+......+(1)/(log_(1983)n)` is equal to

Options

A. `-1`
B. `0`
C. `1`
D. `2`

Answer

To solve the expression

\[ \frac{1}{\log_2 n} + \frac{1}{\log_3 n} + \frac{1}{\log_4 n} + \ldots + \frac{1}{\log_{1983} n} \]

where \( n = 1983! \), we can use the change of base formula for logarithms. The change of base formula states that

\[ \log_b a = \frac{\log_k a}{\log_k b} \]

for any positive \( k \). We will use base 10 for simplicity.

### Step 1: Rewrite the logarithms using the change of base formula

Using the change of base formula, we can rewrite each term in the expression:

\[ \log_k n = \frac{\log n}{\log k} \]

Thus, we have:

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

### Step 2: Substitute into the expression

Now substituting this back into our original expression gives us:

\[ \frac{1}{\log_2 n} + \frac{1}{\log_3 n} + \ldots + \frac{1}{\log_{1983} n} = \frac{\log 2}{\log n} + \frac{\log 3}{\log n} + \ldots + \frac{\log 1983}{\log n} \]

### Step 3: Factor out \(\frac{1}{\log n}\)

We can factor out \(\frac{1}{\log n}\):

\[ = \frac{1}{\log n} \left( \log 2 + \log 3 + \log 4 + \ldots + \log 1983 \right) \]

### Step 4: Use the property of logarithms

Using the property of logarithms that states \(\log a + \log b = \log(ab)\), we can combine the logs:

\[ \log 2 + \log 3 + \log 4 + \ldots + \log 1983 = \log(2 \times 3 \times 4 \times \ldots \times 1983) \]

This product is equal to \(\log(1983!)\).

### Step 5: Substitute back into the expression

Now substituting this back into our expression gives us:

\[ = \frac{\log(1983!)}{\log n} \]

Since \( n = 1983! \), we have:

\[ = \frac{\log(1983!)}{\log(1983!)} = 1 \]

### Final Answer

Thus, the value of the expression

\[ \frac{1}{\log_2 n} + \frac{1}{\log_3 n} + \frac{1}{\log_4 n} + \ldots + \frac{1}{\log_{1983} n} = 1. \]

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