The sequence `{x_(k)}` is defined by `x_(k+1)=x_(k)^(2)+x_(k)` and `x_(1)=(1)/(2)`. Then `[(1)/(x_(1)+1)+(1)/(x_(2)+1)+...+(1)/(x_(100)+1)]` (where `[.]` denotes the greatest integer function) is equal to

Options

A. `0`
B. `2`
C. `4`
D. `1`

Explanation

To solve the problem step by step, we will analyze the sequence defined and the expression we need to evaluate.

### Step 1: Understand the sequence
The sequence is defined by:
\[ x_{k+1} = x_k^2 + x_k \]
with the initial condition:
\[ x_1 = \frac{1}{2} \]

### Step 2: Calculate the first few terms of the sequence
Let's compute the first few terms of the sequence to identify a pattern.

- For \( k = 1 \):
\[ x_2 = x_1^2 + x_1 = \left(\frac{1}{2}\right)^2 + \frac{1}{2} = \frac{1}{4} + \frac{1}{2} = \frac{1}{4} + \frac{2}{4} = \frac{3}{4} \]

- For \( k = 2 \):
\[ x_3 = x_2^2 + x_2 = \left(\frac{3}{4}\right)^2 + \frac{3}{4} = \frac{9}{16} + \frac{12}{16} = \frac{21}{16} \]

- For \( k = 3 \):
\[ x_4 = x_3^2 + x_3 = \left(\frac{21}{16}\right)^2 + \frac{21}{16} = \frac{441}{256} + \frac{336}{256} = \frac{777}{256} \]

### Step 3: Generalize the expression
Now we need to evaluate:
\[ S = \left(\frac{1}{x_1 + 1} + \frac{1}{x_2 + 1} + \cdots + \frac{1}{x_{100} + 1}\right) \]

### Step 4: Simplify each term
We can rewrite each term:
\[ \frac{1}{x_k + 1} = \frac{1}{x_k} - \frac{1}{x_{k+1}} \]

This means:
\[ S = \left(\frac{1}{x_1} - \frac{1}{x_2}\right) + \left(\frac{1}{x_2} - \frac{1}{x_3}\right) + \cdots + \left(\frac{1}{x_{100}} - \frac{1}{x_{101}}\right) \]

### Step 5: Notice the telescoping nature
The series telescopes, leading to:
\[ S = \frac{1}{x_1} - \frac{1}{x_{101}} \]

### Step 6: Calculate \( \frac{1}{x_1} \) and \( \frac{1}{x_{101}} \)
From our earlier calculations:
\[ x_1 = \frac{1}{2} \Rightarrow \frac{1}{x_1} = 2 \]

As \( k \) increases, \( x_k \) grows larger. We can see that \( x_k \) is increasing and will approach a value greater than 1. Thus, \( \frac{1}{x_{101}} \) will be a small positive number.

### Step 7: Estimate \( S \)
Since \( \frac{1}{x_{101}} \) is small, we can approximate:
\[ S \approx 2 - \text{(small positive number)} \]

### Step 8: Apply the greatest integer function
Since \( S \) is slightly less than 2, we find:
\[ \lfloor S \rfloor = 1 \]

### Final Answer
Thus, the value of the expression is:
\[ \lfloor S \rfloor = 1 \]

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