If n is a positive integer, then `(sqrt3+1)^(2n)- (sqrt3-1)^(2n)` is

Options

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

Answer

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

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

3. **Combine the Expansions**:
Now, we subtract the second expansion from the first:
\[ (\sqrt{3}+1)^{2n} - (\sqrt{3}-1)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} (\sqrt{3})^{2n-k} (1 - (-1)^k) \]

4. **Simplify the Expression**:
Notice that \(1 - (-1)^k\) is:
- \(2\) if \(k\) is odd (since \(1 - (-1) = 2\))
- \(0\) if \(k\) is even (since \(1 - 1 = 0\))

Therefore, the sum only includes terms where \(k\) is odd:
\[ = 2 \sum_{k \text{ odd}} \binom{2n}{k} (\sqrt{3})^{2n-k} \]

5. **Recognize the Result**:
The remaining sum is a combination of terms that will yield an integer multiplied by \(\sqrt{3}\). Thus, the entire expression simplifies to:
\[ 2 \cdot \text{(integer)} \cdot \sqrt{3} \]

6. **Conclusion**:
Since \(2 \cdot \text{(integer)}\) is an integer, we conclude that:
\[ (\sqrt{3}+1)^{2n} - (\sqrt{3}-1)^{2n} = \text{(integer)} \cdot \sqrt{3} \]
This means the expression is an irrational number.

### Final Answer:
Thus, if \(n\) is a positive integer, then \((\sqrt{3}+1)^{2n} - (\sqrt{3}-1)^{2n}\) is an **irrational number**.