`""^((2n + 1))C_0 + ""^((2n+ 1))C_1 + ""^((2n + 1))C_2 + ……+""^((2n + 1))C_n = `

Options

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

Answer

To solve the problem \( \binom{2n + 1}{0} + \binom{2n + 1}{1} + \binom{2n + 1}{2} + \ldots + \binom{2n + 1}{n} \), we can use the Binomial Theorem and some properties of binomial coefficients.

### Step-by-Step Solution:

1. **Understanding the Binomial Theorem**:
The Binomial Theorem states that:
\[ (1 + x)^{n} = \sum_{k=0}^{n} \binom{n}{k} x^k \]
For our case, we will consider \( n = 2n + 1 \).

2. **Setting Up the Equation**:
We can express the sum of the binomial coefficients as:
\[ \sum_{k=0}^{2n+1} \binom{2n+1}{k} = (1 + 1)^{2n+1} = 2^{2n+1} \]
This represents the sum of all coefficients from \( k = 0 \) to \( k = 2n + 1 \).

3. **Using Symmetry of Binomial Coefficients**:
The binomial coefficients have a symmetry property:
\[ \binom{n}{k} = \binom{n}{n-k} \]
Therefore, we can pair the coefficients:
\[ \binom{2n + 1}{0} + \binom{2n + 1}{1} + \ldots + \binom{2n + 1}{n} = \binom{2n + 1}{n} + \binom{2n + 1}{n+1} + \ldots + \binom{2n + 1}{2n + 1} \]

4. **Dividing the Total Sum**:
Since the total sum of coefficients is \( 2^{2n + 1} \), and we know that the coefficients from \( k = 0 \) to \( k = n \) and from \( k = n + 1 \) to \( k = 2n + 1 \) are equal, we can conclude:
\[ \sum_{k=0}^{n} \binom{2n + 1}{k} = \sum_{k=n+1}^{2n+1} \binom{2n + 1}{k} = \frac{1}{2} \cdot 2^{2n + 1} = 2^{2n} \]

5. **Final Result**:
Therefore, we have:
\[ \binom{2n + 1}{0} + \binom{2n + 1}{1} + \binom{2n + 1}{2} + \ldots + \binom{2n + 1}{n} = 2^{2n} \]

### Conclusion:
The final answer is:
\[ \boxed{2^{2n}} \]