Let `alpha_(n) =(2+sqrt(3))^(n)`. Find `lim_(n to oo) (alpha_(n)-[alpha_(n)])`([.] denotes greateset integer function)

Options

A. 1
B. `(1)/(2)`
C. `(1)/(3)`
D. `(2)/(3)`

Answer

To solve the problem, we need to find the limit:

\[ \lim_{n \to \infty} \left( \alpha_n - [\alpha_n] \right) \]

where \(\alpha_n = (2 + \sqrt{3})^n\) and \([\cdot]\) denotes the greatest integer function.

### Step-by-step Solution:

1. **Define the terms**:
Let \(\alpha_n = (2 + \sqrt{3})^n\) and denote the integral part of \(\alpha_n\) as \([ \alpha_n ]\). The fractional part can be expressed as:
\[ \{ \alpha_n \} = \alpha_n - [\alpha_n] \]

2. **Consider the conjugate**:
We also consider the conjugate of \(\alpha_n\):
\[ \beta_n = (2 - \sqrt{3})^n \]
Note that \(2 - \sqrt{3} \approx 0.268\), which is less than 1. Therefore, as \(n\) approaches infinity, \(\beta_n\) approaches 0.

3. **Express the sum**:
Now, we can express the sum of \(\alpha_n\) and \(\beta_n\):
\[ \alpha_n + \beta_n = (2 + \sqrt{3})^n + (2 - \sqrt{3})^n \]
This sum is an integer because it can be derived from the binomial expansion of both terms.

4. **Identify the integral and fractional parts**:
Let:
\[ \alpha_n = i + f \quad \text{and} \quad \beta_n = g \]
where \(i\) is the integral part of \(\alpha_n\), \(f\) is the fractional part of \(\alpha_n\), and \(g\) is the fractional part of \(\beta_n\). Since \(\beta_n\) approaches 0, \(g\) will also be a small positive number.

5. **Combine the parts**:
From the previous step, we have:
\[ i + f + g = \text{integer} \]
Since \(f\) and \(g\) are both in the range \([0, 1)\), the sum \(f + g\) must also be an integer. The only possible integer value for \(f + g\) in the range \([0, 2)\) is 1. Therefore:
\[ f + g = 1 \]

6. **Isolate the fractional part**:
Rearranging gives us:
\[ f = 1 - g \]
As \(n\) approaches infinity, since \(\beta_n \to 0\), \(g\) approaches 0. Hence, \(f\) approaches 1.

7. **Final limit**:
Therefore, we conclude that:
\[ \lim_{n \to \infty} \{ \alpha_n \} = \lim_{n \to \infty} f = 1 \]

### Conclusion:
The final result is:
\[ \lim_{n \to \infty} \left( \alpha_n - [\alpha_n] \right) = 1 \]