If `C_0, C_1,C_2 ..., C_n`, denote the binomial coefficients in the expansion of `(1 + x)^n`, then `C_1/2+C_3/4+C_5/6+......` is equal to

Explanation

To solve the problem, we need to find the value of the sum \( \frac{C_1}{2} + \frac{C_3}{4} + \frac{C_5}{6} + \ldots \) where \( C_k \) are the binomial coefficients from the expansion of \( (1 + x)^n \).

### Step-by-Step Solution:

1. **Write the Binomial Expansion**:
The binomial expansion of \( (1 + x)^n \) is given by:
\[ (1 + x)^n = C_0 + C_1 x + C_2 x^2 + C_3 x^3 + \ldots + C_n x^n \]
where \( C_k = \binom{n}{k} \).

2. **Integrate Both Sides**:
We will integrate both sides with respect to \( x \):
\[ \int (1 + x)^n \, dx = \int (C_0 + C_1 x + C_2 x^2 + C_3 x^3 + \ldots + C_n x^n) \, dx \]
The left side becomes:
\[ \frac{(1 + x)^{n+1}}{n + 1} + C \]
The right side becomes:
\[ C_0 x + \frac{C_1 x^2}{2} + \frac{C_2 x^3}{3} + \frac{C_3 x^4}{4} + \ldots + \frac{C_n x^{n+1}}{n + 1} \]

3. **Evaluate at \( x = 1 \)**:
Substituting \( x = 1 \):
\[ \frac{(1 + 1)^{n + 1}}{n + 1} = \frac{2^{n + 1}}{n + 1} \]
The right side becomes:
\[ C_0 + \frac{C_1}{2} + \frac{C_2}{3} + \frac{C_3}{4} + \ldots + \frac{C_n}{n + 1} \]

4. **Evaluate at \( x = -1 \)**:
Now substituting \( x = -1 \):
\[ \frac{(1 - 1)^{n + 1}}{n + 1} = 0 \]
The right side becomes:
\[ C_0 - \frac{C_1}{2} + \frac{C_2}{3} - \frac{C_3}{4} + \ldots + (-1)^n \frac{C_n}{n + 1} \]

5. **Add the Two Results**:
Now, we add the results from steps 3 and 4:
\[ \frac{2^{n + 1}}{n + 1} + 0 = C_0 + \frac{C_1}{2} + \frac{C_2}{3} + \frac{C_3}{4} + \ldots + \frac{C_n}{n + 1} + C_0 - \frac{C_1}{2} + \frac{C_2}{3} - \frac{C_3}{4} + \ldots + (-1)^n \frac{C_n}{n + 1} \]
The terms involving \( C_1, C_3, \ldots \) cancel out, leading to:
\[ \frac{2^{n + 1}}{n + 1} = 2 \left( \frac{C_1}{2} + \frac{C_3}{4} + \frac{C_5}{6} + \ldots \right) \]

6. **Solve for the Desired Sum**:
Dividing both sides by 2 gives:
\[ \frac{2^{n}}{n + 1} = \frac{C_1}{2} + \frac{C_3}{4} + \frac{C_5}{6} + \ldots \]

Thus, the final answer is:
\[ \frac{2^n}{n + 1} \]