The coeffiecent of `n^(-r)` in the expansion of `log_(10)((n)/(n-1))` is

Options

A. `(1)/(rlog_(e)10)`
B. `-(1)/(rlog_(e)10)`
C. `(1)/(r!log_(e)10)`
D. `log_(e)1-(1)(n)/(log_(e))`

Answer

To find the coefficient of \( n^{-r} \) in the expansion of \( \log_{10} \left( \frac{n}{n-1} \right) \), we can follow these steps:

### Step 1: Rewrite the logarithm
We start with the expression:
\[ \log_{10} \left( \frac{n}{n-1} \right) = \log_{10}(n) - \log_{10}(n-1) \]

### Step 2: Use the logarithm property
We can express \( \log_{10}(n-1) \) using the Taylor series expansion around \( n \) (or \( n \to \infty \)):
\[ \log_{10}(n-1) = \log_{10} \left( n \left(1 - \frac{1}{n}\right) \right) = \log_{10}(n) + \log_{10} \left(1 - \frac{1}{n}\right) \]
Using the logarithmic expansion:
\[ \log_{10}(1 - x) \approx -x - \frac{x^2}{2} - \frac{x^3}{3} - \ldots \quad \text{for small } x \]
we substitute \( x = \frac{1}{n} \):
\[ \log_{10}(1 - \frac{1}{n}) \approx -\frac{1}{n} - \frac{1}{2n^2} - \frac{1}{3n^3} - \ldots \]

### Step 3: Substitute back into the equation
Substituting back into our expression:
\[ \log_{10} \left( \frac{n}{n-1} \right) = \log_{10}(n) - \left( \log_{10}(n) - \left( -\frac{1}{n} - \frac{1}{2n^2} - \frac{1}{3n^3} - \ldots \right) \right) \]
This simplifies to:
\[ \log_{10} \left( \frac{n}{n-1} \right) \approx \frac{1}{n} + \frac{1}{2n^2} + \frac{1}{3n^3} + \ldots \]

### Step 4: Identify the coefficient of \( n^{-r} \)
From the series expansion, we can see that the coefficient of \( n^{-r} \) corresponds to the term \( \frac{1}{r} \) in the series:
\[ \text{Coefficient of } n^{-r} = \frac{1}{r} \]

### Step 5: Include the logarithmic factor
Since we are interested in the coefficient in terms of \( \log_{10}(e) \):
\[ \text{Coefficient} = \frac{1}{r} \cdot \log_{10}(e) \]

### Final Answer
Thus, the coefficient of \( n^{-r} \) in the expansion of \( \log_{10} \left( \frac{n}{n-1} \right) \) is:
\[ \frac{\log_{10}(e)}{r} \]