If n is a positive integer, then `(sqrt(3)+1)^(2n)-(sqrt(3)-1)^(2n)` is (1) an irrational number (2) an odd positive integer (3) an even positive integer (4) a rational number other than positive integers

Options

A. an irrational number
B. an odd positive integer
C. an even positive integer
D. a rational number other than positiveintegers

Explanation

To solve the problem, we need to evaluate the expression \((\sqrt{3}+1)^{2n} - (\sqrt{3}-1)^{2n}\) for a positive integer \(n\).

### Step-by-step Solution:

1. **Rewrite the Expression**:
We start with the expression:
\[ (\sqrt{3}+1)^{2n} - (\sqrt{3}-1)^{2n} \]

2. **Use Binomial Theorem**:
According to the Binomial Theorem, we can expand both terms:
\[ (\sqrt{3}+1)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} (\sqrt{3})^k (1)^{2n-k} \]
\[ (\sqrt{3}-1)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} (\sqrt{3})^k (-1)^{2n-k} \]

3. **Combine the Expansions**:
When we subtract the two expansions, we notice that the terms where \(k\) is odd will cancel out:
\[ (\sqrt{3}+1)^{2n} - (\sqrt{3}-1)^{2n} = 2 \sum_{k \text{ even}} \binom{2n}{k} (\sqrt{3})^k \]
This is because the odd indexed terms will have opposite signs and thus cancel each other.

4. **Factor Out Common Terms**:
We can factor out \(2\) from the even indexed terms:
\[ = 2 \left( \sum_{j=0}^{n} \binom{2n}{2j} (\sqrt{3})^{2j} \right) \]
where \(j\) runs from \(0\) to \(n\).

5. **Simplify the Expression**:
Notice that \((\sqrt{3})^{2j} = 3^j\), so we can rewrite the expression as:
\[ = 2 \sum_{j=0}^{n} \binom{2n}{2j} 3^j \]

6. **Recognize the Result**:
The sum \(\sum_{j=0}^{n} \binom{2n}{2j} 3^j\) is a polynomial in \(3\) with integer coefficients. Therefore, the entire expression is an integer multiplied by \(2\).

7. **Conclusion**:
Since \(2\) times an integer is an even integer, we conclude that:
\[ (\sqrt{3}+1)^{2n} - (\sqrt{3}-1)^{2n} \text{ is an even positive integer.} \]

### Final Answer:
Thus, the correct option is:
(3) an even positive integer.