Sum of n terms of the series
`log a + log((a^2)/(b)) + log((a^3)/(b^2)) + log ((a^4)/(b^3)) + ...` is :

Options

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

Answer

To find the sum of the first \( n \) terms of the series

\[ \log a + \log\left(\frac{a^2}{b}\right) + \log\left(\frac{a^3}{b^2}\right) + \log\left(\frac{a^4}{b^3}\right) + \ldots \]

we can break down the terms step by step.

### Step 1: Rewrite the Terms
The series can be rewritten using logarithmic properties. Recall that:

\[ \log\left(\frac{x}{y}\right) = \log x - \log y \]

Thus, we can express each term in the series as:

\[ \log a + \left(\log a^2 - \log b\right) + \left(\log a^3 - \log b^2\right) + \left(\log a^4 - \log b^3\right) + \ldots \]

### Step 2: Expand the Series
Now, let's expand the series:

\[ \log a + \log a^2 - \log b + \log a^3 - \log b^2 + \log a^4 - \log b^3 + \ldots \]

Grouping the terms gives:

\[ (\log a + \log a^2 + \log a^3 + \log a^4 + \ldots) - (\log b + \log b^2 + \log b^3 + \ldots) \]

### Step 3: Sum the Logarithms of \( a \)
The first part is a sum of logarithms of powers of \( a \):

\[ \log a + \log a^2 + \log a^3 + \ldots + \log a^n = \log(a^1 \cdot a^2 \cdot a^3 \cdots a^n) \]

The product can be simplified using the property of exponents:

\[ = \log(a^{1 + 2 + 3 + \ldots + n}) = \log(a^{\frac{n(n+1)}{2}}) \]

### Step 4: Sum the Logarithms of \( b \)
Now, for the second part:

\[ \log b + \log b^2 + \log b^3 + \ldots + \log b^{n-1} = \log(b^1 \cdot b^2 \cdots b^{n-1}) = \log(b^{1 + 2 + \ldots + (n-1)}) = \log(b^{\frac{(n-1)n}{2}}) \]

### Step 5: Combine the Results
Now, combining both parts, we get:

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

Using the property of logarithms:

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

### Final Step: Simplify
Thus, the sum of the first \( n \) terms of the series is:

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

### Final Answer
The sum of \( n \) terms of the series is:

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