If `x=1+1/(3+1/(3+1/(2...oo)))` then the value of x is

Options

A. `sqrt(5/2)`
B. `sqrt(3/2)`
C. `sqrt(7/3)`
D. `sqrt(5/3)`

Explanation

To solve the equation \( x = 1 + \frac{1}{3 + \frac{1}{3 + \frac{1}{2 + \ldots}}} \), we will follow these steps:

### Step 1: Define the Infinite Expression
Let \( y = 3 + \frac{1}{3 + \frac{1}{2 + \ldots}} \). Then, we can express \( x \) as:
\[ x = 1 + \frac{1}{y} \]

### Step 2: Substitute \( y \) Back into the Equation
Notice that the expression for \( y \) is also recursive. We can express \( y \) in terms of itself:
\[ y = 3 + \frac{1}{y} \]

### Step 3: Rearranging the Equation for \( y \)
Multiply both sides of the equation by \( y \) to eliminate the fraction:
\[ y^2 = 3y + 1 \]

### Step 4: Rearranging to Form a Quadratic Equation
Rearranging gives us:
\[ y^2 - 3y - 1 = 0 \]

### Step 5: Solve the Quadratic Equation
Now, we can use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -3, c = -1 \):
\[ y = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
\[ y = \frac{3 \pm \sqrt{9 + 4}}{2} \]
\[ y = \frac{3 \pm \sqrt{13}}{2} \]

### Step 6: Choose the Positive Root
Since \( y \) must be positive, we take:
\[ y = \frac{3 + \sqrt{13}}{2} \]

### Step 7: Substitute \( y \) Back to Find \( x \)
Now substitute \( y \) back into the equation for \( x \):
\[ x = 1 + \frac{1}{y} = 1 + \frac{2}{3 + \sqrt{13}} \]

### Step 8: Rationalize the Denominator
To simplify \( \frac{2}{3 + \sqrt{13}} \), we multiply the numerator and denominator by the conjugate \( 3 - \sqrt{13} \):
\[ x = 1 + \frac{2(3 - \sqrt{13})}{(3 + \sqrt{13})(3 - \sqrt{13})} \]
\[ = 1 + \frac{2(3 - \sqrt{13})}{9 - 13} = 1 + \frac{2(3 - \sqrt{13})}{-4} \]
\[ = 1 - \frac{1}{2}(3 - \sqrt{13}) = 1 - \frac{3}{2} + \frac{\sqrt{13}}{2} \]
\[ = -\frac{1}{2} + \frac{\sqrt{13}}{2} \]
\[ = \frac{\sqrt{13} - 1}{2} \]

### Final Answer
Thus, the value of \( x \) is:
\[ \boxed{\frac{\sqrt{13} - 1}{2}} \]