Evaluate the `""^12C_0 + ""^12C_2 + ""^12C_4 +………….+""^12C_12`

Answer

To evaluate the expression \( \binom{12}{0} + \binom{12}{2} + \binom{12}{4} + \ldots + \binom{12}{12} \), we can use the Binomial Theorem and properties of binomial coefficients.

### Step-by-Step Solution:

1. **Understand the Binomial Theorem**:
The Binomial Theorem states that:
\[ (x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \]
For our case, we will use \( n = 12 \).

2. **Set Up the Expression**:
We want to evaluate the sum of the even-indexed binomial coefficients from \( \binom{12}{0} \) to \( \binom{12}{12} \). This can be represented as:
\[ S = \binom{12}{0} + \binom{12}{2} + \binom{12}{4} + \ldots + \binom{12}{12} \]

3. **Use the Binomial Theorem**:
We can evaluate \( (1 + 1)^{12} \) and \( (1 - 1)^{12} \):
\[ (1 + 1)^{12} = 2^{12} = \sum_{k=0}^{12} \binom{12}{k} \]
\[ (1 - 1)^{12} = 0^{12} = 0 = \sum_{k=0}^{12} \binom{12}{k} (-1)^k \]

4. **Combine the Two Results**:
The first equation gives us the total sum of all binomial coefficients, while the second gives us the alternating sum:
\[ \sum_{k=0}^{12} \binom{12}{k} (-1)^k = \binom{12}{0} - \binom{12}{1} + \binom{12}{2} - \binom{12}{3} + \ldots + \binom{12}{12} = 0 \]

5. **Add the Two Equations**:
Now, if we add the two results:
\[ 2^{12} + 0 = \sum_{k=0}^{12} \binom{12}{k} + \sum_{k=0}^{12} \binom{12}{k} (-1)^k \]
The odd-indexed terms will cancel out, leaving us with:
\[ 2^{12} = 2 \left( \binom{12}{0} + \binom{12}{2} + \binom{12}{4} + \ldots + \binom{12}{12} \right) \]

6. **Solve for \( S \)**:
Thus, we have:
\[ 2^{12} = 2S \]
Dividing both sides by 2:
\[ S = \frac{2^{12}}{2} = 2^{11} \]

7. **Final Answer**:
Therefore, the value of the expression \( \binom{12}{0} + \binom{12}{2} + \binom{12}{4} + \ldots + \binom{12}{12} \) is:
\[ \boxed{2048} \]