The value of `sum_(r=0)^(3n-1)(-1)^r ^(6n)C_(2r+1)3^r` is

Options

A. `2^(3n)`
B. `2^(2n-1)`
C. `2^(6n-1)`
D. `0`

Explanation

To solve the problem:

**Question:** Find the value of the summation \( \sum_{r=0}^{3n-1} (-1)^r \binom{6n}{2r+1} 3^r \).

### Step-by-step Solution:

1. **Rewrite the Summation:**
We start with the given summation:
\[ S = \sum_{r=0}^{3n-1} (-1)^r \binom{6n}{2r+1} 3^r \]

2. **Use Complex Numbers:**
We can express \( (-1)^r \) in terms of complex exponentials. We know that:
\[ (-1)^r = \left( \frac{1}{\sqrt{3}} i \right)^{2r} \cdot \left( \sqrt{3} i \right)^{2r + 1} \]
Thus, we can rewrite the summation as:
\[ S = \sum_{r=0}^{3n-1} \frac{1}{\sqrt{3} i} \binom{6n}{2r+1} \left( \sqrt{3} i \right)^{2r + 1} \]

3. **Factor Out the Constant:**
The constant \( \frac{1}{\sqrt{3} i} \) can be factored out of the summation:
\[ S = \frac{1}{\sqrt{3} i} \sum_{r=0}^{3n-1} \binom{6n}{2r+1} \left( \sqrt{3} i \right)^{2r + 1} \]

4. **Recognize the Binomial Expansion:**
The summation can be interpreted as the imaginary part of the expansion of \( (1 + \sqrt{3} i)^{6n} \):
\[ (1 + \sqrt{3} i)^{6n} = 2^{6n} \left( \cos(6n \cdot \frac{\pi}{3}) + i \sin(6n \cdot \frac{\pi}{3}) \right) \]

5. **Evaluate the Cosine and Sine:**
We know that:
\[ \cos(6n \cdot \frac{\pi}{3}) = \cos(2n\pi) = 1 \]
and
\[ \sin(6n \cdot \frac{\pi}{3}) = \sin(2n\pi) = 0 \]

6. **Combine Results:**
Therefore, the imaginary part of \( (1 + \sqrt{3} i)^{6n} \) is:
\[ \text{Imaginary part} = 0 \]

7. **Final Result:**
Putting it all together, we find:
\[ S = \frac{1}{\sqrt{3} i} \cdot 0 = 0 \]

### Conclusion:
The value of the summation \( \sum_{r=0}^{3n-1} (-1)^r \binom{6n}{2r+1} 3^r \) is \( \boxed{0} \).