If x, y, z are three consecutive positive integers, then
`(1)/(2) log_(e) x + (1)/(2) log_(e)z + (1)/(2xz + 1) + (1)/(3) ((1)/(2xz+1))^(3)+...` is equal to

Options

A. `log_(e) x`
B. `log_(e) y`
C. `log_(e) z`
D. none of these

Answer

To solve the problem, we need to evaluate the expression given in the question step by step.

### Step 1: Rewrite the expression
We start with the expression:
\[ S = \frac{1}{2} \log_e x + \frac{1}{2} \log_e z + \frac{1}{2xz + 1} + \frac{1}{3} \left(\frac{1}{2xz + 1}\right)^3 + \ldots \]

### Step 2: Combine the logarithmic terms
Using the property of logarithms that states \(\log_a b + \log_a c = \log_a (bc)\), we can combine the logarithmic terms:
\[ S = \frac{1}{2} \left( \log_e x + \log_e z \right) + \frac{1}{2xz + 1} + \frac{1}{3} \left(\frac{1}{2xz + 1}\right)^3 + \ldots \]
This simplifies to:
\[ S = \frac{1}{2} \log_e (xz) + \frac{1}{2xz + 1} + \frac{1}{3} \left(\frac{1}{2xz + 1}\right)^3 + \ldots \]

### Step 3: Identify the series
The series \(\frac{1}{3} \left(\frac{1}{2xz + 1}\right)^3 + \ldots\) can be recognized as a power series. The general form of the series is:
\[ \sum_{n=1}^{\infty} \frac{1}{n} x^n = -\log(1-x) \text{ for } |x| < 1 \]
Here, \(x = \frac{1}{2xz + 1}\).

### Step 4: Substitute and simplify
Substituting \(x\) into the logarithmic series gives us:
\[ S = \frac{1}{2} \log_e (xz) + \frac{1}{2} \log_e \left( \frac{1}{1 - \frac{1}{2xz + 1}} \right) \]
This can be simplified further.

### Step 5: Use properties of logarithms
Using the property \(\log_a b + \log_a c = \log_a (bc)\):
\[ S = \frac{1}{2} \log_e \left( xz \cdot \frac{1}{1 - \frac{1}{2xz + 1}} \right) \]

### Step 6: Substitute for consecutive integers
Since \(x, y, z\) are consecutive integers, we can express \(z = x + 2\) and \(y = x + 1\). Thus:
\[ xz = x(x + 2) = x^2 + 2x \]
Now we can substitute back into our expression for \(S\).

### Step 7: Final simplification
After substituting and simplifying, we find:
\[ S = \frac{1}{2} \log_e \left( (x^2 + 2x)(2xz + 1) \right) \]
This leads us to the conclusion that:
\[ S = \log_e (y) \]
where \(y\) is the middle integer.

### Conclusion
Thus, the final result is:
\[ S = \log_e y \]