The sum `S_(n)=sum_(k=0)^(n)(-1)^(k)*^(3n)C_(k)`, where `n=1,2,….` is

Options

A. `(-1)^(n)*"^(3n-1)C_(n-1)`
B. `(-1)^(n)*"^(3n-1)C_(n)`
C. `(-1)n*"^(3n-1)C_(n+1)`
D. None of these

Explanation

To solve the problem, we need to evaluate the sum

\[ S_n = \sum_{k=0}^{n} (-1)^k \binom{3n}{k} \]

where \( n = 1, 2, \ldots \).

### Step-by-Step Solution:

1. **Understanding the Sum**:
The sum can be expressed as:
\[ S_n = \binom{3n}{0} - \binom{3n}{1} + \binom{3n}{2} - \binom{3n}{3} + \ldots + (-1)^n \binom{3n}{n} \]

2. **Using the Binomial Theorem**:
The Binomial Theorem states that:
\[ (1 + x)^m = \sum_{k=0}^{m} \binom{m}{k} x^k \]
By substituting \( x = -1 \), we get:
\[ (1 - 1)^{3n} = \sum_{k=0}^{3n} \binom{3n}{k} (-1)^k = 0 \]

3. **Splitting the Sum**:
We can separate the terms in the sum:
\[ 0 = \sum_{k=0}^{3n} \binom{3n}{k} (-1)^k = \sum_{k=0}^{n} \binom{3n}{k} (-1)^k + \sum_{k=n+1}^{3n} \binom{3n}{k} (-1)^k \]
The first sum is \( S_n \) and the second sum can be rewritten using the property of binomial coefficients.

4. **Using the Identity**:
We can use the identity:
\[ \binom{m}{k} = \binom{m-1}{k-1} + \binom{m-1}{k} \]
This allows us to express \( S_n \) in terms of smaller binomial coefficients.

5. **Final Expression**:
After simplification, we find that:
\[ S_n = (-1)^n \binom{3n-1}{n} \]

### Final Result:
Thus, the sum \( S_n \) can be expressed as:
\[ S_n = (-1)^n \binom{3n-1}{n} \]