If ` (x + a_(1)) (x + a_(2)) (x + a_(3)) …(x + a_(n)) = x^(n) + S_(1) x^(n-1) + S_(2) x^(n-2) + …+ S_(n)`
where , ` S_(1) = sum_(i=0)^(n) a_(i), S_(2) = (sumsum)_(1lei lt j le n) a_(i) a_(j) , S_(3) (sumsumsum)_(1le i ltk le n) a_(i) a_(j) a_(k)`
and so on .
If `(1 + x)^(n) = C_(0) + C_(1) x + C_(2)x^(2) + ...+ C_(n) x^(n)` the
cefficient of ` x^(n)` in the expansion of
` (x + C_(0))(x + C_(1)) (x + C_(2))...(x + C_(n)) ` is

Options

A. `2^(2n-1) - (1)/(2) ""^(2n)C_(n-1)`
B. `2^(2n-1) - (1)/(2) ""^(2n)C_(n)`
C. `2^(2n-1) - (1)/(2) ""^(2n+1)C_(n)`
D. `2^(2n-1) - (1)/(2) ""^(2n+1)C_(n-1)`

Explanation

To solve the problem, we need to find the coefficient of \( x^n \) in the expansion of \( (x + C_0)(x + C_1)(x + C_2) \ldots (x + C_n) \), where \( C_k \) are the coefficients from the binomial expansion of \( (1 + x)^n \).

### Step-by-Step Solution:

1. **Understand the Expression**:
We start with the expression:
\[ (x + C_0)(x + C_1)(x + C_2) \ldots (x + C_n) \]
This is a product of \( n+1 \) terms.

2. **Identify the Coefficient of \( x^n \)**:
The coefficient of \( x^n \) in this expansion can be found by selecting \( x \) from \( n \) of the \( (x + C_k) \) terms and \( C_k \) from one term.

3. **Calculate the Coefficient**:
The coefficient of \( x^n \) will be the sum of all possible products of \( C_k \) taken one at a time:
\[ C_0 + C_1 + C_2 + \ldots + C_n \]

4. **Use the Binomial Theorem**:
From the binomial theorem, we know that:
\[ (1 + x)^n = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n \]
Setting \( x = 1 \) gives:
\[ (1 + 1)^n = 2^n = C_0 + C_1 + C_2 + \ldots + C_n \]
Therefore, the sum \( C_0 + C_1 + C_2 + \ldots + C_n = 2^n \).

5. **Final Result**:
Thus, the coefficient of \( x^n \) in the expansion of \( (x + C_0)(x + C_1)(x + C_2) \ldots (x + C_n) \) is:
\[ 2^n \]

### Summary:
The coefficient of \( x^n \) in the expansion of \( (x + C_0)(x + C_1)(x + C_2) \ldots (x + C_n) \) is \( 2^n \).