The value of `sum_(n=0)^(m) log ((a^(2n-1))/(b^(m-1)))(a != 0 , 1 , b != 0 , 1 )` is

Options

A. `m log. (a^(2m))/(b^(m-1))`
B. `(m+1)log. a/(b^(m-1))`
C. `m/2 log -(a^(2m))/(b^(2m-2))`
D. `m/2 log. (a^(2m))/(b^(m+1))`

Answer

To solve the given problem, we need to evaluate the sum:

\[ S = \sum_{n=0}^{m} \log \left( \frac{a^{2n-1}}{b^{m-1}} \right) \]

### Step-by-Step Solution:

1. **Rewrite the Summation:**
We can break down the logarithm using the properties of logarithms:
\[ S = \sum_{n=0}^{m} \left( \log(a^{2n-1}) - \log(b^{m-1}) \right) \]

2. **Separate the Terms:**
The sum can be separated into two parts:
\[ S = \sum_{n=0}^{m} \log(a^{2n-1}) - \sum_{n=0}^{m} \log(b^{m-1}) \]

3. **Simplify the Second Term:**
Since \(\log(b^{m-1})\) does not depend on \(n\), it can be factored out:
\[ S = \sum_{n=0}^{m} \log(a^{2n-1}) - (m+1) \log(b^{m-1}) \]

4. **Evaluate the First Summation:**
The first summation can be simplified further:
\[ S = \sum_{n=0}^{m} (2n-1) \log(a) - (m+1) \log(b^{m-1}) \]

This can be rewritten as:
\[ S = \log(a) \sum_{n=0}^{m} (2n-1) - (m+1) \log(b^{m-1}) \]

5. **Calculate the Summation of \(2n-1\):**
The summation \(\sum_{n=0}^{m} (2n-1)\) can be calculated as follows:
\[ \sum_{n=0}^{m} (2n-1) = 2\sum_{n=0}^{m} n - \sum_{n=0}^{m} 1 = 2 \cdot \frac{m(m+1)}{2} - (m+1) = m(m+1) - (m+1) = m^2 \]

6. **Substitute Back:**
Now substituting back into the equation:
\[ S = \log(a) m^2 - (m+1) \log(b^{m-1}) \]

7. **Final Simplification:**
We can express the final result as:
\[ S = m^2 \log(a) - (m+1)(m-1) \log(b) \]

### Final Result:
Thus, the value of the summation is:
\[ S = m^2 \log(a) - (m+1)(m-1) \log(b) \]