`C_0+C_1+C_2+.........+C_(N-1)+C_N=`

Options

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

Answer

To solve the problem \( C_0 + C_1 + C_2 + \ldots + C_{N-1} + C_N \), where \( C_k \) represents the binomial coefficient \( \binom{N}{k} \), we can use the Binomial Theorem.

### Step-by-Step Solution:

1. **Understanding the Binomial Theorem**:
The Binomial Theorem states that:
\[ (x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \]
where \( \binom{n}{k} \) is the binomial coefficient.

2. **Setting Values for \( x \) and \( y \)**:
To find the sum of the binomial coefficients from \( C_0 \) to \( C_N \), we can set \( x = 1 \) and \( y = 1 \):
\[ (1 + 1)^N = \sum_{k=0}^{N} \binom{N}{k} 1^{N-k} 1^k \]

3. **Calculating the Left Side**:
The left side simplifies to:
\[ 2^N \]

4. **Writing the Right Side**:
The right side becomes:
\[ \sum_{k=0}^{N} \binom{N}{k} = \binom{N}{0} + \binom{N}{1} + \binom{N}{2} + \ldots + \binom{N}{N} \]

5. **Conclusion**:
Therefore, we have:
\[ C_0 + C_1 + C_2 + \ldots + C_N = 2^N \]

### Final Answer:
\[ C_0 + C_1 + C_2 + \ldots + C_N = 2^N \]