Find the sum of 'n' terms of the series.
`log_(2)(x/y) + log_(4)(x/y)^(2) + log_(8)(x/y)^(3) + log_(16)(x/y)^(4)` +……..

Options

A. `log_(2)(x/y)^(4n)`
B. `n (log_(2)(x/y))`
C. `log_(2)(x^(n-1)/y^(n-1))`
D. `1/2 log_(2)(x/y)^(n(n+1))`

Answer

To find the sum of the series given by

\[ S_n = \log_{2}(x/y) + \log_{4}((x/y)^{2}) + \log_{8}((x/y)^{3}) + \log_{16}((x/y)^{4}) + \ldots \]

we will analyze each term in the series and then sum them up.

### Step 1: Rewrite each term using logarithmic properties

We know that the logarithmic property states:

\[ \log_{b}(a^n) = n \cdot \log_{b}(a) \]

Using this property, we can rewrite the terms in the series:

1. The first term is \(\log_{2}(x/y)\).
2. The second term can be rewritten as:
\[ \log_{4}((x/y)^{2}) = 2 \cdot \log_{4}(x/y) \]
Since \(4 = 2^2\), we can convert the base:
\[ \log_{4}(x/y) = \frac{1}{2} \cdot \log_{2}(x/y) \implies 2 \cdot \log_{4}(x/y) = 2 \cdot \frac{1}{2} \cdot \log_{2}(x/y) = \log_{2}(x/y) \]

3. The third term:
\[ \log_{8}((x/y)^{3}) = 3 \cdot \log_{8}(x/y) \]
Since \(8 = 2^3\), we convert the base:
\[ \log_{8}(x/y) = \frac{1}{3} \cdot \log_{2}(x/y) \implies 3 \cdot \log_{8}(x/y) = 3 \cdot \frac{1}{3} \cdot \log_{2}(x/y) = \log_{2}(x/y) \]

4. The fourth term:
\[ \log_{16}((x/y)^{4}) = 4 \cdot \log_{16}(x/y) \]
Since \(16 = 2^4\), we convert the base:
\[ \log_{16}(x/y) = \frac{1}{4} \cdot \log_{2}(x/y) \implies 4 \cdot \log_{16}(x/y) = 4 \cdot \frac{1}{4} \cdot \log_{2}(x/y) = \log_{2}(x/y) \]

### Step 2: Summing up the terms

Now, we can see that each term simplifies to \(\log_{2}(x/y)\):

\[ S_n = \log_{2}(x/y) + \log_{2}(x/y) + \log_{2}(x/y) + \ldots \text{ (n terms)} \]

Thus, the sum of \(n\) terms is:

\[ S_n = n \cdot \log_{2}(x/y) \]

### Final Answer

The sum of the first \(n\) terms of the series is:

\[ S_n = n \cdot \log_{2}(x/y) \]