Find the sum to n terms of the series
`log a+ log"" (a^3)/(b) + log ""(a^5)/(b^2) + log ""(a^7)/(b^3)+ `….

Options

A. `log((a^(2n))/(b^(n-1)))^(n//2)`
B. `log"" (a^(2n-1))/(b^(n-1))`
C. `log""((a^(2n))/(b^n))`
D. none of these

Answer

To find the sum to n terms of the series:

\[ S_n = \log a + \log \frac{a^3}{b} + \log \frac{a^5}{b^2} + \log \frac{a^7}{b^3} + \ldots \]

we can break down the series into two parts: the logarithms of the numerators and the logarithms of the denominators.

### Step 1: Rewrite the series
We can express the series as:

\[ S_n = \left( \log a + \log a^3 + \log a^5 + \ldots + \log a^{(2n-1)} \right) - \left( \log b + \log b^2 + \log b^3 + \ldots + \log b^{(n-1)} \right) \]

### Step 2: Simplify the first part
The first part of the series can be simplified using the property of logarithms:

\[ \log a + \log a^3 + \log a^5 + \ldots + \log a^{(2n-1)} = \log \left( a^{1 + 3 + 5 + \ldots + (2n-1)} \right) \]

The sum of the first n odd numbers is \( n^2 \). Therefore, we have:

\[ 1 + 3 + 5 + \ldots + (2n-1) = n^2 \]

Thus, the first part becomes:

\[ \log \left( a^{n^2} \right) = n^2 \log a \]

### Step 3: Simplify the second part
The second part of the series is:

\[ \log b + \log b^2 + \log b^3 + \ldots + \log b^{(n-1)} = \log \left( b^{1 + 2 + 3 + \ldots + (n-1)} \right) \]

The sum of the first \( n-1 \) natural numbers is given by:

\[ 1 + 2 + 3 + \ldots + (n-1) = \frac{(n-1)n}{2} \]

Thus, the second part becomes:

\[ \log \left( b^{\frac{(n-1)n}{2}} \right) = \frac{(n-1)n}{2} \log b \]

### Step 4: Combine both parts
Now, we can combine both parts to find the sum \( S_n \):

\[ S_n = n^2 \log a - \frac{(n-1)n}{2} \log b \]

### Final Result
Therefore, the sum to n terms of the series is:

\[ S_n = n^2 \log a - \frac{(n-1)n}{2} \log b \]